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Find points on a curve where the gradient equals a given value

A gradient condition is an equation in x

Differentiate $y = f (x)$ to get its gradient $f^{'} (x)$ . Solve $f^{'} (x) = m$ for the required gradient $m$ . Each resulting $x$ is only a horizontal coordinate; substitute it into the original $f (x)$ to get the full point.

Problem

The curve is $y = \frac{1}{3} x^{3} - \frac{11}{2} x^{2} + 24 x$ . Find the coordinates of the two points where the gradient is $- 4$ . Round decimal coordinates to one decimal place.

Answer

(4, 29.3)

and

(7, 12.8)

Exact y-coordinates are 88/3 and 77/6, respectively.

Step-by-step solution

1. Differentiate

\frac{d y}{d x} = x^{2} - 11 x + 24.

2. Set the gradient to -4

x^{2} - 11 x + 24 = - 4 \Rightarrow x^{2} - 11 x + 28 = 0.

Factor: $(x - 4) (x - 7) = 0$ , so $x = 4$ or $x = 7$ .

3. Find the points

Substitute into the original cubic, not into its derivative:

y (4) = \frac{64}{3} - 88 + 96 = \frac{88}{3} \approx 29.3,

y (7) = \frac{343}{3} - \frac{539}{2} + 168 = \frac{77}{6} \approx 12.8.

Therefore the two points are $(4, 29.3)$ and $(7, 12.8)$ to one decimal place.

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