Increasing and Decreasing Curves

Video lesson · from 49:112 worked examples

One lesson video covers all of Differentiation, so it opens at 49:11 for this topic — not from the beginning.

If $y$ increases as $x$ increases, the curve is strictly increasing.

Tangents slope upwards, so their gradients are positive: $\frac{dy}{dx} > 0$ .

If $y$ decreases as $x$ increases, the curve is strictly decreasing.

Tangents slope downwards, so their gradients are negative: $\frac{dy}{dx} < 0$ .

Answer with an inequality, not a number: the question asks for the values of $x$ for which the function is increasing, so the answer is a range such as $x \gt 2$ .
Justify with the sign of the derivative: state $\frac{dy}{dx} \gt 0$ for increasing and $\frac{dy}{dx} \lt 0$ for decreasing. An unsupported answer does not gain full marks.
"Show that it is always increasing" needs an argument: usually complete the square on the derivative to show it can never be negative. Testing a few values proves nothing.
Strict inequalities at stationary points: where $\frac{dy}{dx} = 0$ the function is neither increasing nor decreasing.

Example 1

Show that the function $f(x) = x^3 - 3x^2 + 3x - 10$ is never decreasing.

To show it is never decreasing, we must show $f'(x) \geq 0$ for all $x$ .

\begin{aligned} f'(x) &= 3x^2 - 6x + 3 \\ &= 3(x^2 - 2x + 1) \\ &= 3(x - 1)^2 \end{aligned}

Since $(x - 1)^2 \geq 0$ for all real $x$ , then $3(x-1)^2 \geq 0$ .

Therefore, $f'(x) \geq 0$ , meaning the curve is never decreasing.

Example 2

Show that the curve with equation $y = 5 - 4x - x^3$ is always decreasing.

To show it is always decreasing, we must show $\frac{dy}{dx} < 0$ for all $x$ .

\begin{aligned} \frac{dy}{dx} &= -4 - 3x^2 \\ &= - (4 + 3x^2) \end{aligned}

Since $3x^2 \geq 0$ for all real $x$ , then $4 + 3x^2 > 0$ .

So $-(4 + 3x^2) < 0$ .

Therefore, $\frac{dy}{dx} < 0$ , meaning the curve is always decreasing.