7.5 Tangents, normals, maxima, minima and stationary points — Practise

Work through these on your own. Find the stationary points before you classify them, and check that every answer makes sense in the context.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Find the value of $x$ at the stationary point of $y = x^2 - 8x + 3$ .
Find the two values of $x$ at the stationary points of $y = 2x^3 - 6x$ .
The curve $y = x^3 - 3x^2 + 2$ has stationary points at $x = 0$ and $x = 2$ . Use the second derivative $6x - 6$ to classify them.
Find the gradient of the tangent to $y = x^2 + 3x$ at $x = 1$ .
Find the gradient of the normal to $y = x^2$ at $x = 1$ .
Find the gradient of the tangent to $y = x^3 - 3x$ at $x = 2$ .
The volume of an open box is $V = x(16 - 2x)(10 - 2x)$ . Find the two values of $x$ for which $\dfrac{dV}{dx} = 12x^2 - 104x + 160$ is zero.
For the box with $V = 4x^3 - 52x^2 + 160x$ , work out the value of $\dfrac{d^2V}{dx^2}$ at $x = 2$ .
A rectangle has perimeter $20$ cm. Its area is $A = x(10 - x)$ for a side $x$ . Find the value of $x$ that gives the greatest area.
Find the value of $x$ at the stationary point of $y = e^x - x$ .
A ball is thrown so that its height after $t$ seconds is $h = 20t - 5t^2$ metres. Find the time at which the height is greatest.
Find the value of $x$ at the stationary point of $y = x^3$ , and state the gradient there.

Answers and concise solutions

1. The value of $x$

Answer: $\dfrac{dy}{dx} = 2x - 8$ , so $2x - 8 = 0$ gives $x = 4$ . The second derivative is $2 > 0$ , so this is a minimum.

2. The two values of $x$

Answer: $\dfrac{dy}{dx} = 6x^2 - 6 = 6(x^2 - 1)$ , so $x^2 = 1$ and $x = 1$ or $x = -1$ .

3. The sign at each point

Answer: At $x = 0$ the second derivative is $-6 < 0$ , so the point is a maximum. At $x = 2$ it is $6 > 0$ , so the point is a minimum.

4. The gradient

Answer: $\dfrac{dy}{dx} = 2x + 3$ . At $x = 1$ the gradient is $2 + 3 = 5$ . The tangent through $(1, 4)$ is $y = 5x - 1$ .

5. The gradient

Answer: The tangent gradient is $2x = 2$ at $x = 1$ . The normal gradient is the negative reciprocal, $-\dfrac{1}{2}$ . The normal through $(1, 1)$ is $y = -\tfrac{1}{2}x + \tfrac{3}{2}$ .

6. The gradient

Answer: $\dfrac{dy}{dx} = 3x^2 - 3$ . At $x = 2$ the gradient is $12 - 3 = 9$ . The tangent through $(2, 2)$ is $y = 9x - 16$ .

7. The two values of $x$

Answer: Divide by $4$ : $3x^2 - 26x + 40 = 0$ . Factorise as $(3x - 20)(x - 2) = 0$ , so $x = \dfrac{20}{3}$ or $x = 2$ . Only $x = 2$ fits the card, because the other value makes the base wider than $10$ cm.

8. The second derivative

Answer: $\dfrac{d^2V}{dx^2} = 24x - 104$ . At $x = 2$ : $48 - 104 = -56 < 0$ , so the volume is a maximum there.

9. The value of $x$

Answer: $\dfrac{dA}{dx} = 10 - 2x = 0$ gives $x = 5$ . The second derivative is $-2 < 0$ , so the area is a maximum. The area is $5 \times 5 = 25$ cm $^2$ .

10. The value of $x$

Answer: $\dfrac{dy}{dx} = e^x - 1 = 0$ , so $e^x = 1$ and $x = 0$ . The second derivative is $e^x$ , which is $1 > 0$ at $x = 0$ , so the point is a minimum.

11. The time

Answer: $\dfrac{dh}{dt} = 20 - 10t = 0$ , so $t = 2$ seconds. The second derivative is $-10 < 0$ , so the height is a maximum. The greatest height is $h(2) = 20$ m.

12. The value of $x$ and the gradient

Answer: $\dfrac{dy}{dx} = 3x^2 = 0$ at $x = 0$ , and the gradient there is $0$ . The curve flattens at the origin but keeps rising, so this is a stationary point of inflection.