Question

The tangent to the curve $y=x^{3}-6 x^{2}+9 x+4,0 \leq x \leq 5$ has maximum slope at $x=k$ then find value of $k$.

Answer 1

$\frac{d y}{d x}=3 x^{2}-12 x+9$
Let $u=3 x^{2}-12 x+9$
$\frac{d u}{d x}=6 x-12=0$
$\Rightarrow x=2$
at $x=0\,\, u=9$
$x=2\,\, u=-3$
$x=5\,\, u=24$
So $\frac{d y}{d x}$ is maximum at $x=5$.