Max/Min, Increasing/Decreasing & Sketching

Section 1 of 3

Maximum and Minimum Points

At the top of a hill or the bottom of a valley, the curve momentarily stops rising or falling — its slope is zero. These are the turning points.

Turning points

Set $\dfrac{dy}{dx} = 0$ and solve. The second derivative tells you which: $\dfrac{d^2y}{dx^2} > 0 \Rightarrow$ minimum (U-shape); $\dfrac{d^2y}{dx^2} < 0 \Rightarrow$ maximum (n-shape).

Worked example — a minimum

Find the minimum point of $y = x^2 - 6x + 3$.

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

$y = 3^2 - 6(3) + 3 = -6$

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

Minimum $(3, -6)$

Worked example — a maximum

Find the maximum point of $y = 7 - 6x - x^2$.

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

$y = 7 - 6(-3) - (-3)^2 = 16$

$\dfrac{d^2y}{dx^2} = -2 < 0 \Rightarrow$ maximum

Maximum $(-3, 16)$

You try

Find the turning point of $y = x^2 - 3x + 1$ and state its nature.

$\tfrac{dy}{dx}=2x-3=0$. Solve, sub back, then check the sign of $\tfrac{d^2y}{dx^2}$.

$2x - 3 = 0 \Rightarrow x = 1.5$

$y = (1.5)^2 - 3(1.5) + 1 = -1.25$

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

Minimum $(1.5, -1.25)$

You try

$y = x^2 + ax + b$ has a turning point at $(2, 6)$. Find $a$ and $b$.

$\tfrac{dy}{dx}=2x+a=0$ at $x=2$ gives $a$. Then use $y=6$ at $x=2$ for $b$.

$2(2) + a = 0 \Rightarrow a = -4$

$2^2 - 4(2) + b = 6$

$4 - 8 + b = 6$

$a = -4,\ b = 10$

Section 2 of 3

Increasing and Decreasing

A curve is increasing where its slope is positive and decreasing where its slope is negative. Form $\dfrac{dy}{dx}$ and solve the inequality.

The test

Increasing: $\dfrac{dy}{dx} > 0$. Decreasing: $\dfrac{dy}{dx} < 0$.

Worked example — increasing

For what values of $x$ is $y = x^2 - 6x + 1$ increasing?

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

$2x > 6$

$x > 3$

Worked example — decreasing

For what values of $x$ is $y = x^2 - 10x + 1$ decreasing?

$\dfrac{dy}{dx} = 2x - 10 < 0$

$2x < 10$

$x < 5$

You try

For what values of $x$ is $y = x^2 - 7x - 1$ increasing?

Set $\tfrac{dy}{dx}=2x-7 > 0$ and solve.

$2x - 7 > 0$

$2x > 7$

$x > \tfrac{7}{2}$

Section 3 of 3

Roots, Turning Points & Sketching

Know where a curve crosses the $x$-axis (its roots) and where it turns, and you can sketch it. Mark the roots and the turning point on the axes — the shape draws itself.

The two ingredients

Roots (curve cuts the $x$-axis): set $y = 0$ and solve, $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Turning point: set $\dfrac{dy}{dx} = 0$.

Worked example — find the minimum, the roots, then sketch

Given $y = x^2 - 6x - 1$, find the minimum point and the roots, then sketch.

$\dfrac{dy}{dx} = 2x - 6 = 0 \Rightarrow x = 3$, $y = 3^2 - 6(3) - 1 = -10$

Roots ($a=1,b=-6,c=-1$): $x = \dfrac{6 \pm \sqrt{36 + 4}}{2} \approx 6.2$ or $-0.2$

Minimum $(3, -10)$; roots $x \approx -0.2$ and $6.2$

Worked example — factorising for the roots

$f(x) = x^2 - 2x - 8$. Find the minimum point and the roots.

$f'(x) = 2x - 2 = 0 \Rightarrow x = 1$, $f(1) = -9$

Roots: $x^2 - 2x - 8 = (x - 4)(x + 2) \Rightarrow x = 4$ or $x = -2$

Min $(1, -9)$; roots $x = -2, 4$

You try

$f(x) = 8 - 4x - x^2$. Find the maximum point and the roots.

$f'(x) = -4 - 2x = 0$ for the turning point. Roots from the formula, $a=-1, b=-4, c=8$.

$f'(x) = -4 - 2x = 0 \Rightarrow x = -2$, $f(-2) = 12$

Roots: $x = \dfrac{4 \pm \sqrt{16 + 32}}{-2} \approx 1.5$ or $-5.5$

Max $(-2, 12)$; roots $x \approx -5.5, 1.5$

That’s Class 3.

Turning points and their nature, increasing/decreasing, and sketching from roots and turning points. Class 4: distance, speed and acceleration.