Inc/Dec & Turning Points

Section 1 of 13

Increasing & Decreasing

As you move left-to-right along a curve, it's either going up, going down, or momentarily flat. Slope tells us which.

Increasing & Decreasing

Increasing $\Longleftrightarrow\;$ $\dfrac{dy}{dx} > 0$

Decreasing $\Longleftrightarrow\;$ $\dfrac{dy}{dx} < 0$

Two flavours of question come from this:

Type A "Show that $y = \ldots$ is always increasing/decreasing." — prove a sign holds everywhere.

Type B "For what $x$ is $y = \ldots$ increasing/decreasing?" — solve an inequality.

Section 2 of 13

Always increasing or decreasing?

Is $\;y = \dfrac{5x - 1}{x - 7}\;$ increasing? ($x \in \mathbb{R}$, $x$ ≠ $7$)

Quotient rule: $u = 5x - 1, \;\; v = x - 7$.

$\dfrac{du}{dx} = 5 \;\;\; \dfrac{dv}{dx} = 1$

$\dfrac{dy}{dx} = \dfrac{5(x - 7) - 1(5x - 1)}{(x - 7)^{2}}$

$\quad\;\; = \dfrac{5x - 35 - 5x + 1}{(x - 7)^{2}}$

$\quad\;\; = \dfrac{-34}{(x - 7)^{2}}$

Now read the signs of numerator and denominator:

Top: $\;-34 < 0$

Bottom: $\;(x - 7)^{2} > 0 \;$ (square is positive)

$\dfrac{dy}{dx} < 0 \;\;\Rightarrow\;\;$ Decreasing (not increasing)

Key move: don't try to compute the sign at one $x$ — argue from the form. A negative number divided by a positive number is always negative.

You try

Show that $y = \dfrac{2x + 3}{x - 1}$ is always decreasing on its domain ($x$ ≠ $1$).

Quotient rule, then read the signs of the top and bottom.

$u = 2x + 3, \;\; v = x - 1$

$\dfrac{du}{dx} = 2, \;\; \dfrac{dv}{dx} = 1$

$\dfrac{dy}{dx} = \dfrac{2(x - 1) - 1(2x + 3)}{(x - 1)^{2}} = \dfrac{-5}{(x - 1)^{2}}$

Top $-5 < 0$; bottom $(x-1)^{2} > 0$ (square is positive).

$\dfrac{dy}{dx} < 0 \;\Rightarrow\;$ always decreasing.

$\dfrac{dy}{dx} = \dfrac{-5}{(x-1)^{2}} < 0$, so always decreasing.

You try

Show that $y = \dfrac{x + 4}{x - 2}$ is always decreasing on its domain ($x$ ≠ $2$).

Same routine — quotient rule, then read the signs.

$u = x + 4, \;\; v = x - 2$

$\dfrac{du}{dx} = 1, \;\; \dfrac{dv}{dx} = 1$

$\dfrac{dy}{dx} = \dfrac{1(x - 2) - 1(x + 4)}{(x - 2)^{2}} = \dfrac{-6}{(x - 2)^{2}}$

Top $-6 < 0$; bottom $(x-2)^{2} > 0$ (square is positive).

$\dfrac{dy}{dx} < 0 \;\Rightarrow\;$ always decreasing.

$\dfrac{dy}{dx} = \dfrac{-6}{(x-2)^{2}} < 0$, so always decreasing.

Section 3 of 13

Range of $x$ for increasing/decreasing

For these, you solve an inequality in the derivative.

For what $x$ is $\;y = x^{2} - 6x + 1\;$ decreasing?

$\dfrac{dy}{dx} = 2x - 6 < 0$

$2x < 6$

$x < 3$

Check against the graph: $\;y = (x - 3)^{2} - 8\;$ is a U-curve with minimum at $x = 3$. To the left of $x = 3$ it's coming down — decreasing. To the right, going up.

For what $x$ is $\;y = x^{3} - 3x^{2} + 1\;$ increasing?

$\dfrac{dy}{dx} = 3x^{2} - 6x > 0$

$x^{2} - 2x > 0$ (divide by 3)

Solve the equation $\;x^{2} - 2x = 0\;$ to find the boundary points:

$x(x - 2) = 0$

$x = 0 \;$ or $\; x = 2$

A positive quadratic ($\cup$-shaped) is positive outside its roots:

$x < 0 \;$ or $\; x > 2$

For what $x$ is $\;y = x^{3} - 3x^{2} - 9x + 1\;$ decreasing?

$\dfrac{dy}{dx} = 3x^{2} - 6x - 9 < 0$

$x^{2} - 2x - 3 < 0$ (divide by 3)

$x^{2} - 2x - 3 = 0$

$(x - 3)(x + 1) = 0$

$x = 3 \;$ or $\; x = -1$

A positive quadratic is negative between its roots:

$-1 < x < 3$

You try

For what values of $x$ is $y = x^{2} + 4x - 7$ increasing?

Set $\dfrac{dy}{dx} > 0$ and solve.

$\dfrac{dy}{dx} = 2x + 4 > 0$

$2x > -4$

$x > -2$

You try

For what values of $x$ is $y = x^{3} - 12x + 5$ decreasing?

$\dfrac{dy}{dx} = 3x^{2} - 12 < 0$. Solve the equation first, then read the sign.

$\dfrac{dy}{dx} = 3x^{2} - 12 < 0$

$x^{2} - 4 < 0$

$x^{2} - 4 = 0 \;\;\Rightarrow\;\; x = \pm 2$

Positive quadratic negative between roots:

$-2 < x < 2$

You try

For what values of $x$ is $y = x^{3} - 3x^{2} - 24x + 1$ increasing?

$\dfrac{dy}{dx} = 3x^{2} - 6x - 24 > 0$. Divide by 3, factor, then think about sign.

$\dfrac{dy}{dx} = 3x^{2} - 6x - 24 > 0$

$x^{2} - 2x - 8 > 0$

$x^{2} - 2x - 8 = 0$

$(x - 4)(x + 2) = 0$

$x = 4 \;$ or $\; x = -2$

Positive quadratic positive outside roots:

$x < -2 \;$ or $\; x > 4$

Section 4 of 13

Turning Points

A turning point is where the curve briefly stops going up (or down) and switches direction. Locally a high (maximum) or a low (minimum).

Also called stationary points.

Right at the top of a hump, or right at the bottom of a dip, the tangent is horizontal. So the slope is zero:

Turning point

$\dfrac{dy}{dx} = 0$

That gives the $x$-value(s). To find $y$, sub back into the original curve. But how do we tell whether it's a max or a min?

Section 5 of 13

The second-derivative test

Differentiate again. The sign of $\dfrac{d^{2}y}{dx^{2}}$ at the turning point tells you the nature:

Nature of a turning point

$\dfrac{d^{2}y}{dx^{2}} > 0 \;\;\Longrightarrow\;\;$ minimum

$\dfrac{d^{2}y}{dx^{2}} < 0 \;\;\Longrightarrow\;\;$ maximum

Intuition: at a minimum the slope is going from negative to positive (slope is increasing) — so the rate of change of the slope is positive. At a maximum, slope is decreasing — second derivative negative.

Section 6 of 13

Find the turning point and state its nature

Find the turning point of $\;y = x^{2} - 6x + 1\;$ and state its nature.

$y = x^{2} - 6x + 1$

$\dfrac{dy}{dx} = 2x - 6 = 0$

$x = 3$

Sub into the original to get $y$:

$y = (3)^{2} - 6(3) + 1 = 9 - 18 + 1 = -8$

Turning point: $(3, -8)$

Test the nature:

$\dfrac{d^{2}y}{dx^{2}} = 2 > 0 \;\;\Rightarrow\;\; $ minimum

Minimum at $(3, -8)$

You try

Find the turning point of $y = x^{2} + 4x - 5$ and state its nature.

Set $\dfrac{dy}{dx} = 0$ to find $x$, sub in for $y$, then check $\dfrac{d^{2}y}{dx^{2}}$.

$\dfrac{dy}{dx} = 2x + 4 = 0 \;\;\Rightarrow\;\; x = -2$

$y = (-2)^{2} + 4(-2) - 5 = 4 - 8 - 5 = -9$

$\dfrac{d^{2}y}{dx^{2}} = 2 > 0 \;\Rightarrow\;$ minimum

Minimum at $(-2, -9)$

You try

Find the turning points of $y = x^{3} - 3x^{2} - 9x + 5$ and state the nature of each.

A cubic has two turning points. Factor the derivative to get them. Then test each with $\dfrac{d^{2}y}{dx^{2}}$.

$\dfrac{dy}{dx} = 3x^{2} - 6x - 9 = 0$

$x^{2} - 2x - 3 = 0$

$(x - 3)(x + 1) = 0$

$x = 3$ or $x = -1$

$y(3) = 27 - 27 - 27 + 5 = -22$

$y(-1) = -1 - 3 + 9 + 5 = 10$

$\dfrac{d^{2}y}{dx^{2}} = 6x - 6$

At $x = 3:\;\; 6(3) - 6 = 12 > 0 \;\Rightarrow\;$ minimum

At $x = -1:\;\; 6(-1) - 6 = -12 < 0 \;\Rightarrow\;$ maximum

Max at $(-1, 10)$, Min at $(3, -22)$

Section 7 of 13

Harder: $f(x) = \dfrac{\ln x}{x}$

A classic exam-style turning-point problem with a clever deduction at the end.

$f(x) = \dfrac{\ln x}{x},\;\; x > 0$.

(i) Show the maximum of $f(x)$ occurs at the point $\left(e, \dfrac{1}{e}\right)$.

(ii) Hence, show that $\;x^{e} \le e^{x}\;$ for all $x > 0$.

Part (i) Quotient rule:

$u = \ln x, \;\; v = x$

$\dfrac{du}{dx} = \dfrac{1}{x}, \;\; \dfrac{dv}{dx} = 1$

$\dfrac{dy}{dx} = \dfrac{v \, \dfrac{du}{dx} - u \, \dfrac{dv}{dx}}{v^{2}} = \dfrac{x \cdot \dfrac{1}{x} - \ln x}{x^{2}} = \dfrac{1 - \ln x}{x^{2}}$

Set the numerator $= 0$ (a fraction is zero only when its top is zero):

$1 - \ln x = 0$

$\ln x = 1$

Recall: $\;\ln x = \log_{e} x\;$. So $\;\log_{e} x = 1 \;\Longleftrightarrow\; e^{1} = x$:

$x = e$

Sub back to find $y$:

$y = \dfrac{\ln e}{e} = \dfrac{1}{e}$ (since $\ln e = 1$)

Turning point: $\left(e, \dfrac{1}{e}\right)$

Now confirm it's a maximum using the second derivative. Quotient rule again on $\;\dfrac{1 - \ln x}{x^{2}}$:

$u = 1 - \ln x, \;\; v = x^{2}$

$\dfrac{du}{dx} = -\dfrac{1}{x}, \;\; \dfrac{dv}{dx} = 2x$

$\dfrac{d^{2}y}{dx^{2}} = \dfrac{x^{2}\left(-\dfrac{1}{x}\right) - 2x(1 - \ln x)}{(x^{2})^{2}}$

$\quad\;\; = \dfrac{-x - 2x(1 - \ln x)}{x^{4}}$

$\quad\;\; = \dfrac{-x - 2x + 2x \ln x}{x^{4}} = \dfrac{-3x + 2x \ln x}{x^{4}}$

$\quad\;\; = \dfrac{-3 + 2 \ln x}{x^{3}}$ (divide top and bottom by $x$)

At $x = e$, $\;\ln e = 1$:

$\dfrac{d^{2}y}{dx^{2}}\Big|_{x=e} = \dfrac{-3 + 2}{e^{3}} = -\dfrac{1}{e^{3}} < 0 \;\;\Rightarrow\;\;$ maximum

Maximum at $\left(e, \dfrac{1}{e}\right) \quad \checkmark$

Part (ii) Deduction.

"Hence" means: use what you just proved. We know $f(x) = \dfrac{\ln x}{x}$ has a maximum value of $\dfrac{1}{e}$, so for all $x > 0$:

$\dfrac{\ln x}{x} \le \dfrac{1}{e}$

Multiply across by $ex$ (which is positive since $x > 0$):

$e \ln x \le x$

$x = x \cdot 1 = x \ln e$, so:

$e \ln x \le x \ln e$

$\ln x^{e} \le \ln e^{x}$ (power rule for logs both sides)

$\ln$ is an increasing function, so we can drop it from both sides (the inequality keeps its direction):

$x^{e} \le e^{x} \quad \checkmark$

You try

$f(x) = \dfrac{\ln x}{x^{2}}, \;\; x > 0$. Find the $x$-coordinate of the turning point and state its nature.

Quotient rule with $u = \ln x, \, v = x^{2}$. Set the numerator $= 0$.

$u = \ln x, \;\; v = x^{2}$

$\dfrac{du}{dx} = \dfrac{1}{x}, \;\; \dfrac{dv}{dx} = 2x$

$\dfrac{dy}{dx} = \dfrac{x^{2}\cdot\dfrac{1}{x} - 2x \ln x}{x^{4}} = \dfrac{x - 2x \ln x}{x^{4}} = \dfrac{1 - 2\ln x}{x^{3}}$

$1 - 2\ln x = 0 \;\Rightarrow\; \ln x = \dfrac{1}{2} \;\Rightarrow\; x = \sqrt{e}$

(Without going through $\dfrac{d^{2}y}{dx^{2}}$, the function $\to -\infty$ as $x \to 0^{+}$ and $\to 0$ as $x \to \infty$, so this single turning point is a maximum.)

Maximum at $x = \sqrt{e}$

Section 8 of 13

Proving there's no maximum

Sometimes a problem asks you to rule out a maximum (or minimum). The trick: show the second derivative has a fixed sign everywhere.

Let $\;g(x) = x^{2} + \dfrac{a}{x^{2}}\;$ where $a$ is a real number and $x$ ≠ $0$. Given $g(x)$ has a turning point at $x = 2$:

(i) find the value of $a$

(ii) prove that $g(x)$ has no local maximum points.

Part (i)

Rewrite with a negative index so we can differentiate cleanly:

$g(x) = x^{2} + ax^{-2}$

$g'(x) = 2x - 2ax^{-3} = 2x - \dfrac{2a}{x^{3}}$

At a turning point, $g'(x) = 0$. Sub $x = 2$:

$g'(2) = 4 - \dfrac{2a}{8} = 0$

$32 - 2a = 0$ (multiply across by 8)

$a = 16$

Part (ii) Prove no maximum.

A maximum needs $g''(x) < 0$. If we can show $g''(x) > 0$ everywhere, no maximum is possible.

$g'(x) = 2x - 32x^{-3}$ (using $a = 16$)

$g''(x) = 2 + 96x^{-4} = 2 + \dfrac{96}{x^{4}}$

Read the signs:

$2 > 0$

$x^{4} = (x^{2})^{2} > 0$ (square is positive)

$\dfrac{96}{x^{4}} > 0$

$g''(x) > 0$ for all $x$ ≠ $0 \;\Rightarrow\;$ no local maximum.

(Any turning point must therefore be a minimum.)

You try

Let $h(x) = x^{2} + \dfrac{16}{x}$, $x$ ≠ $0$. Show that $h''(x)$ is positive for all $x > 0$, and deduce that any turning point of $h$ on the positive $x$-axis must be a minimum.

Write $h(x) = x^{2} + 16x^{-1}$, then differentiate twice. Read signs.

$h(x) = x^{2} + 16x^{-1}$

$h'(x) = 2x - 16x^{-2}$

$h''(x) = 2 + 32x^{-3} = 2 + \dfrac{32}{x^{3}}$

For $x > 0$: $\;2 > 0\;$ and $\;\dfrac{32}{x^{3}} > 0$, so $h''(x) > 0$.

Any turning point in $x > 0$ has $h''(x) > 0 \;\Rightarrow\;$ minimum.

$h''(x) = 2 + \dfrac{32}{x^{3}} > 0$ for $x > 0$, so any turning point there is a minimum.

Section 9 of 13

Show a local max, then deduce something

$f(x) = \log_{e}(3x) - 3x, \;\; x > 0$.

(i) Show that $\left(\dfrac{1}{3}, -1\right)$ is a local maximum point of $f(x)$.

(ii) Deduce that the graph of $f(x)$ does not intersect the $x$-axis.

Part (i)

$f(x) = \ln(3x) - 3x$

Differentiate. The derivative of $\ln(3x)$ uses chain rule: outer is $\ln u$ (gives $\dfrac{1}{u}$), inner is $3x$ (gives $3$):

$f'(x) = \dfrac{3}{3x} - 3 = \dfrac{1}{x} - 3$

Set to zero:

$\dfrac{1}{x} - 3 = 0$

$\dfrac{1}{x} = 3 \;\;\Rightarrow\;\; x = \dfrac{1}{3}$

Sub $x = \dfrac{1}{3}$ into the original:

$f\left(\dfrac{1}{3}\right) = \ln\!\left(3 \cdot \tfrac{1}{3}\right) - 3 \cdot \tfrac{1}{3} = \ln 1 - 1 = -1$ (since $\ln 1 = 0$)

Turning point: $\left(\dfrac{1}{3}, -1\right) \;\checkmark$

Confirm it's a maximum:

$f'(x) = x^{-1} - 3$

$f''(x) = -x^{-2} = -\dfrac{1}{x^{2}}$

$-1 < 0, \;\;\;\; x^{2} > 0 \;$ (square is positive)

$f''(x) < 0 \;\;\Rightarrow\;\;$ maximum

Local maximum at $\left(\dfrac{1}{3}, -1\right) \;\checkmark$

Part (ii) Deduction.

The maximum value of $f$ is $-1$. So the curve never gets higher than $-1$. Since $-1 < 0$, the whole graph sits below the $x$-axis.

Maximum $y$-value is $-1 < 0$, so the curve can never cut the $x$-axis. $\;\checkmark$

You try

$g(x) = \ln(2x) - 2x, \;\; x > 0$. Show that $\left(\dfrac{1}{2}, -1\right)$ is a local maximum and hence deduce that $g(x) \le -1$ for all $x > 0$.

Same routine as the worked example — $g'(x) = \dfrac{1}{x} - 2$.

$g'(x) = \dfrac{2}{2x} - 2 = \dfrac{1}{x} - 2 = 0$

$x = \dfrac{1}{2}$

$g\!\left(\dfrac{1}{2}\right) = \ln 1 - 1 = -1$

$g''(x) = -\dfrac{1}{x^{2}} < 0 \;\Rightarrow\;$ maximum

Max value is $-1$, so $g(x) \le -1$ for all $x > 0$. $\;\checkmark$

Max at $\left(\dfrac{1}{2}, -1\right)$; hence $g(x) \le -1$.

Max at $\left(\dfrac{1}{2}, -1\right)$; hence $g(x) \le -1$ for all $x > 0$.

Section 10 of 13

Cubics & the point of inflection

A cubic typically has two turning points (a max and a min) and one point of inflection sitting between them — where the curve changes from concave-down to concave-up (or vice versa).

Point of inflection

$\dfrac{d^{2}y}{dx^{2}} = 0$

The curve isn't bending — it's about to switch from concave one way to concave the other.

For $\;f(x) = x^{3} - 3x + 5\;$:

(i) find the turning points and classify each

(ii) find the point of inflection.

Part (i) Turning points: $f'(x) = 0$.

$f'(x) = 3x^{2} - 3 = 0$

$x^{2} = 1 \;\;\Rightarrow\;\; x = \pm 1$

$f(1) = 1 - 3 + 5 = 3 \;\;\;\;\;\; f(-1) = -1 + 3 + 5 = 7$

Classify each with $f''(x)$:

$f''(x) = 6x$

$f''(-1) = -6 < 0 \;\Rightarrow\;$ maximum at $(-1, 7)$

$f''(1) = 6 > 0 \;\Rightarrow\;$ minimum at $(1, 3)$

Part (ii) Point of inflection: $f''(x) = 0$.

$6x = 0 \;\;\Rightarrow\;\; x = 0$

$f(0) = 5$

Point of inflection: $(0, 5)$

Sketch check: max at $(-1, 7)$, drop through inflection at $(0, 5)$, down to min at $(1, 3)$, then back up. The min sits at $y = 3 > 0$, so the curve is fully above the $x$-axis to the right of the min — only one real root (somewhere far to the left).

You try

For $f(x) = x^{3} - 12x + 1$, find the turning points (and classify each) and the point of inflection.

$f'(x) = 0$ for the turning points, $f''(x) = 0$ for the inflection.

$f'(x) = 3x^{2} - 12 = 0$

$x^{2} = 4 \;\Rightarrow\; x = \pm 2$

$f(2) = 8 - 24 + 1 = -15$

$f(-2) = -8 + 24 + 1 = 17$

$f''(x) = 6x$

$f''(-2) = -12 < 0 \;\Rightarrow\;$ max at $(-2, 17)$

$f''(2) = 12 > 0 \;\Rightarrow\;$ min at $(2, -15)$

Inflection: $6x = 0 \;\Rightarrow\; x = 0, \; f(0) = 1$

Max $(-2, 17)$, Min $(2, -15)$, Inflection $(0, 1)$

You try

For $f(x) = 2x^{3} - 6x^{2} + 5$, find the turning points (classified) and the point of inflection.

$f'(x) = 6x^{2} - 12x$. Factor out the $6x$.

$f'(x) = 6x^{2} - 12x = 6x(x - 2) = 0$

$x = 0$ or $x = 2$

$f(0) = 5$

$f(2) = 16 - 24 + 5 = -3$

$f''(x) = 12x - 12$

$f''(0) = -12 < 0 \;\Rightarrow\;$ max at $(0, 5)$

$f''(2) = 12 > 0 \;\Rightarrow\;$ min at $(2, -3)$

Inflection: $12x - 12 = 0 \;\Rightarrow\; x = 1, \; f(1) = 2 - 6 + 5 = 1$

Max $(0, 5)$, Min $(2, -3)$, Inflection $(1, 1)$

Section 11 of 13

Finding the coefficients of a cubic

Reverse the problem: you're told properties of a cubic, find the cubic itself. Each property gives one equation.

$f(x) = ax^{3} + bx^{2} + cx + d\;$ has a maximum at $(0, 4)$ and a point of inflection at $(1, 0)$. Find $a, b, c, d$.

Write down everything we can differentiate:

$f(x) = ax^{3} + bx^{2} + cx + d$

$f'(x) = 3ax^{2} + 2bx + c$

$f''(x) = 6ax + 2b$

Translate each given fact into an equation.

Fact 1: Max at $(0, 4) \;\Rightarrow\; f'(0) = 0$.

$f'(0) = 3a(0)^{2} + 2b(0) + c = c = 0$

$c = 0$

Fact 2: The point $(0, 4)$ is on the curve $\;\Rightarrow\; f(0) = 4$.

$f(0) = d = 4$

$d = 4$

Fact 3: Inflection at $(1, 0) \;\Rightarrow\; f''(1) = 0$.

$f''(1) = 6a(1) + 2b = 6a + 2b = 0$

$3a + b = 0$ (divide by 2) ... (i)

Fact 4: The point $(1, 0)$ is on the curve $\;\Rightarrow\; f(1) = 0$.

$f(1) = a + b + c + d = 0$

Sub in $c = 0, \; d = 4$:

$a + b + 0 + 4 = 0$

$a + b = -4$ ... (ii)

Solve the simultaneous equations (i) and (ii). Subtract (ii) from (i):

$(3a + b) - (a + b) = 0 - (-4)$

$2a = 4 \;\;\Rightarrow\;\; a = 2$

Sub back into (ii):

$2 + b = -4 \;\;\Rightarrow\;\; b = -6$

$a = 2, \;\; b = -6, \;\; c = 0, \;\; d = 4$

Check: $\;f(x) = 2x^{3} - 6x^{2} + 4\;$. We've just done this one — max at $(0, 5)$? Wait, let's verify $f(0) = 4 \checkmark$, $f(1) = 2 - 6 + 4 = 0 \checkmark$. Good.

You try

$f(x) = ax^{3} + bx^{2} + cx + d$ has a maximum at $(0, 2)$ and a point of inflection at $(1, 0)$. Find $a, b, c, d$.

Same four facts: $f'(0) = 0, \; f(0) = 2, \; f''(1) = 0, \; f(1) = 0$.

$f'(0) = c = 0$

$f(0) = d = 2$

$f''(1) = 6a + 2b = 0 \;\Rightarrow\; 3a + b = 0$ ... (i)

$f(1) = a + b + c + d = 0 \;\Rightarrow\; a + b = -2$ ... (ii)

(i) $-$ (ii): $\; 2a = 2 \;\Rightarrow\; a = 1$

$b = -3$

$a = 1, \; b = -3, \; c = 0, \; d = 2$

Section 12 of 13

Number of roots of a cubic

Once you know the two turning points of a cubic, you can read off how many times the curve crosses the $x$-axis — its number of real roots. Four scenarios:

Cubic root scenarios

(a) Max above the $x$-axis, min below $\Rightarrow\;$ 3 distinct real roots

(b) One turning point sits exactly on the $x$-axis $\Rightarrow\;$ 3 real roots, two of them equal

(c) Both turning points on the same side of the $x$-axis $\Rightarrow\;$ 1 real root

(d) No turning points at all (derivative has no real roots) $\Rightarrow\;$ 1 real root

Method: find the turning points, look at the signs of their $y$-values, decide which case applies.

How many real roots does $\;f(x) = x^{3} - 3x + 1\;$ have?

$f'(x) = 3x^{2} - 3 = 0 \;\Rightarrow\; x = \pm 1$

$f(1) = 1 - 3 + 1 = -1$

$f(-1) = -1 + 3 + 1 = 3$

$f''(x) = 6x$

$f''(-1) = -6 < 0 \;\Rightarrow\;$ max at $(-1, 3)$

$f''(1) = 6 > 0 \;\Rightarrow\;$ min at $(1, -1)$

Max $y = 3 > 0$, min $y = -1 < 0$ — turning points sit on opposite sides of the $x$-axis. That's case (a):

3 distinct real roots.

You try

How many real roots does $f(x) = x^{3} - 3x + 5$ have?

We found the turning points in section 10: max at $(-1, 7)$, min at $(1, 3)$. Look at the signs.

Max at $(-1, 7)$, min at $(1, 3)$.

Both $y$-values are positive — both turning points sit above the $x$-axis.

That's case (c): both turning points on the same side.

1 real root.

1 real root (both turning points lie above the $x$-axis).

Section 13 of 13

Wrap-up

Three derivative facts pull all this work together:

Quick reference

$\bullet$ $\dfrac{dy}{dx} > 0$ — increasing $\;\;\dfrac{dy}{dx} < 0\;$ — decreasing

$\bullet$ $\dfrac{dy}{dx} = 0$ — turning point

$\bullet$ $\dfrac{d^{2}y}{dx^{2}} > 0$ — minimum $\;\;\dfrac{d^{2}y}{dx^{2}} < 0\;$ — maximum $\;\;\dfrac{d^{2}y}{dx^{2}} = 0\;$ — point of inflection

The "hence" parts at the end of harder questions are usually just applying the max/min value you've just found — bounding the function above or below.

That's Inc/Dec & Turning Points.

Next lesson takes this calculus and applies it to word problems — boxes, sectors, cylinders, rectangles inscribed in shapes. The maths is the same; the work is in setting up the function in the first place.