Question

Maximum slope of the curve $y=-x^3+3 x^2+9 x-27$ is

(a) 0

(b) 12

(c) 16

(d) 32

Answer 1

SOLUTION — Let $y=f(x)=-x^3+3 x^2+9 x-27$

The slope of this curve $f^{\prime}(x)=-3 x^2+6 x+9$

Let

$g(x)=f^{\prime}(x)=-3 x^2+6 x+9$

On differentiating w.r.t. $x$, we get

$g^{\prime}(x)=-6 x+6$

For maxima or minima put $g^{\prime}(x)=0 \Rightarrow x=1$

Now, $g^{\prime \prime}(x)=-6<0$ and hence at $x=1, g(x)$ slope will have maximum value.

$\therefore \quad[g(1)]_{\max }=-3 \times 1+6(1)+9=12$

So, The correct option of this question will be (B).