Question

The points on the curve y = 12x - x³ at which the gradient is zero are _______________.

Answer 1

Let (h, k) be the point on the curve y = 12x − x³ at which the gradient (or slope) of the tangent is zero.

∴ k = 12h − h³ .....(1)

y = 12x − x³

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=12-3x^2$

$\therefore {\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=12-3h^2=0$ (Given)

$\Rightarrow 12-3h^2=0$

$\Rightarrow h^2=4$

⇒ h = −2 or h = 2

Putting h = −2 in (1), we get

k = 12 × (−2) − (−2)³ = −24 + 8 = −16

Putting h = 2 in (1), we get

k = 12 × 2 − (2)³ = 24 − 8 = 16

So, the coordinates of the required point are (−2, −16) and (2, 16).

Thus, the points on the curve y = 12x − x³ at which the gradient is zero are (−2, −16) and (2, 16).


The points on the curve y = 12x − x³ at which the gradient is zero are ___(−2, −16) and (2, 16)___.