Question

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

Answer 1

Answer: 5.00

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