Question

At what point, the slope of the curve $\text{y}=-{\text{x}}^3+3{\text{x}}^2+9\text{x}-27$ is maximum? Also find the maximum slope

Answer 1

Given: $\text{y}=-{\text{x}}^3+3{\text{x}}^2+9\text{x}-27$
Let slope of given curve be denoted by f(x)
$\text{f'(x)}=\frac{\text{dy}}{\text{dx}}=-3{\text{x}}^2+6\text{x}+9$
Critical points of f(x) are given by
$\text{f'(x)}=0$
Now, $\text{f'(x)}=-6\text{x}+6$
$\Rightarrow -6\text{x}+6=0$
$\Rightarrow \text{x}=1$
$\text{f''(x)}=-6$
$\therefore {\text{f''(x)|}}_{\text{x}=1}=-6<0$
Thus, f(x) has maxima when $\text{x}=1$
$\therefore \text{Maximum slope}={\text{f''(x)|}}_{\text{x}=1}$
$=3(1{)}^2+6(1)+9=12$
Substituting $\text{x}=1$ in equation of curve To find the value of 𝑦
$y=-(1{)}^3+3(1{)}^2+9\left(1\right)-27=-16$
Hence, at point $(1,-16)$ the slope of the given curve is maximum and maximum slope is $12$.