Stationary Points (VCE SSCE Mathematical Methods): Revision Notes

Worked Example: Finding Stationary Points of a Quadratic

Find the stationary points of $y = 9 + 12x - 2x^2$.

Solution:

First, find the derivative:

Set the derivative equal to zero:

Now find the $y$-coordinate by substituting $x = 3$ into the original function:

Therefore, the stationary point is $(3, 27)$.

Worked Example: Finding Stationary Points of a Cubic

Find the stationary points of $y = 4 + 3x - x^3$.

Solution:

Find the derivative:

Set equal to zero:

When $x = 1$: $y = 4 + 3(1) - (1)^3 = 6$

When $x = -1$: $y = 4 + 3(-1) - (-1)^3 = 2$

The stationary points are (1, 6) and (-1, 2).

Worked Example: Trigonometric Functions

Find the stationary points of $y = \sin(2x)$ for $x \in [0, 2\pi]$.

Solution:

Find the derivative:

Set equal to zero:

This occurs when:

Therefore:

Substituting these values:

• When $x = \frac{\pi}{4}$: $y = \sin\left(\frac{\pi}{2}\right) = 1$
• When $x = \frac{3\pi}{4}$: $y = \sin\left(\frac{3\pi}{2}\right) = -1$
• When $x = \frac{5\pi}{4}$: $y = \sin\left(\frac{5\pi}{2}\right) = 1$
• When $x = \frac{7\pi}{4}$: $y = \sin\left(\frac{7\pi}{2}\right) = -1$

The stationary points are $\left(\frac{\pi}{4}, 1\right)$, $\left(\frac{3\pi}{4}, -1\right)$, $\left(\frac{5\pi}{4}, 1\right)$, and $\left(\frac{7\pi}{4}, -1\right)$.

Worked Example: Exponential Functions

Find the stationary points of $y = e^{2x} - x$.

Solution:

Find the derivative:

Set equal to zero:

Taking natural logarithms:

When $x = -\frac{1}{2}\ln 2$:

The stationary point is $\left(-\frac{1}{2}\ln 2, \frac{1}{2} + \frac{1}{2}\ln 2\right)$.

Worked Example: Logarithmic Functions

Find the stationary points of $y = x\log_e(2x)$ for $x \in (0, \infty)$.

Solution:

Using the product rule:

Set equal to zero:

When $x = \frac{1}{2e}$:

The stationary point is $\left(\frac{1}{2e}, -\frac{1}{2e}\right)$.

Worked Example: Finding Unknown Parameters

The curve with equation $y = x^3 + ax^2 + bx + c$ passes through the point $(0, 5)$ and has a stationary point at $(2, 7)$. Find $a$, $b$, and $c$.

Solution:

Since the curve passes through $(0, 5)$:

When $x = 0$, $y = 5$, so $c = 5$

Find the derivative:

Since there's a stationary point at $(2, 7)$, we know $\frac{dy}{dx} = 0$ when $x = 2$:

Also, the point $(2, 7)$ lies on the curve:

Multiply equation 2 by 2:

Rearranging equation 1:

Substituting into equation 2:

Therefore: $b = -12 - 4\left(-\frac{9}{2}\right) = -12 + 18 = 6$

The values are: $a = -\frac{9}{2}$, $b = 6$, and $c = 5$.

Worked Example: Using Gradient Tables

For the function $f(x) = 3x^3 - 4x + 1$, find the stationary points and state their nature.

Solution:

Find the derivative:

Find stationary points by solving $f'(x) = 0$:

Calculate the $y$-coordinates:

When $x = -\frac{2}{3}$: $f\left(-\frac{2}{3}\right) = 3\left(-\frac{2}{3}\right)^3 - 4\left(-\frac{2}{3}\right) + 1 = \frac{27}{9} = 3$

When $x = \frac{2}{3}$: $f\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^3 - 4\left(\frac{2}{3}\right) + 1 = -\frac{7}{9}$

The stationary points are $\left(-\frac{2}{3}, 3\right)$ and $\left(\frac{2}{3}, -\frac{7}{9}\right)$.

Now test the sign of $f'(x)$ in each region:

For $x < -\frac{2}{3}$, try $x = -1$: $f'(-1) = 9 - 4 = 5 > 0$ ✓ (positive)

For $-\frac{2}{3} < x < \frac{2}{3}$, try $x = 0$: $f'(0) = -4 < 0$ ✗ (negative)

For $x > \frac{2}{3}$, try $x = 1$: $f'(1) = 9 - 4 = 5 > 0$ ✓ (positive)

Create the gradient table:

Conclusion:

• At $x = -\frac{2}{3}$: gradient changes from $+$ to $-$, so there's a local maximum at $\left(-\frac{2}{3}, 3\right)$
• At $x = \frac{2}{3}$: gradient changes from $-$ to $+$, so there's a local minimum at $\left(\frac{2}{3}, -\frac{7}{9}\right)$

Worked Example: Sketching an Exponential Function

Sketch the graph of $f(x) = e^{x^3}$.

Solution:

End behavior: As $x \to -\infty$, $x^3 \to -\infty$, so $e^{x^3} \to 0$

Axis intercepts: When $x = 0$: $f(0) = e^0 = 1$

The curve passes through $(0, 1)$.

Stationary points:

For stationary points: $3x^2 e^{x^3} = 0$

Since $e^{x^3} > 0$ for all $x$, we need $3x^2 = 0$, giving $x = 0$.

The only stationary point is at $(0, 1)$.

Nature of stationary point: Notice that $f'(x) = 3x^2 e^{x^3} \geq 0$ for all $x$ (it's always positive or zero).

The gradient is never negative, so the point $(0, 1)$ is a stationary point of inflection.

Worked Example: Sketching a Logarithmic Function

For $f:(0, \infty) \to \mathbb{R}$, where $f(x) = x\log_e x$, sketch the graph.

Solution:

Derivative: Using the product rule:

Solving $f(x) = 0$:

Since $x > 0$, we need $\log_e x = 0$, giving $x = 1$.

The curve crosses the $x$-axis at $(1, 0)$.

Stationary points: $f'(x) = 0$ when $\log_e x + 1 = 0$

When $x = e^{-1}$: $y = e^{-1} \times (-1) = -e^{-1}$

Stationary point: $\left(e^{-1}, -e^{-1}\right)$

Nature: For $x < e^{-1}$: $f'(x) < 0$ (gradient negative)

For $x > e^{-1}$: $f'(x) > 0$ (gradient positive)

Therefore, $\left(e^{-1}, -e^{-1}\right)$ is a local minimum.

Worked Example: Multiple Turning Points

Find the local maximum and local minimum points of $f(x) = 2\sin x + 1 - 2\sin^2 x$ where $0 < x < 2\pi$.

Solution:

Find the derivative:

Set equal to zero:

This gives us: $\cos x = 0$ or $\sin x = \frac{1}{2}$

Solutions in the given interval:

From $\cos x = 0$: $x = \frac{\pi}{2}, \frac{3\pi}{2}$

From $\sin x = \frac{1}{2}$: $x = \frac{\pi}{6}, \frac{5\pi}{6}$

Calculate $y$-values:

Create a gradient table:

Conclusions:

• Local maxima at $\left(\frac{\pi}{6}, \frac{3}{2}\right)$ and $\left(\frac{5\pi}{6}, \frac{3}{2}\right)$
• Local minima at $\left(\frac{\pi}{2}, 1\right)$ and $\left(\frac{3\pi}{2}, -3\right)$