Question

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

• A. $0$
• B. $2$
• C. $\frac{5}{2}$
• D. $5$

Answer 1

The correct option is D $5$

$y=x^3-6x^2+9x+4$
$\frac{dy}{dx}=3x^2-12x+9$

Slope of tangent,

$m=f\left(x\right)=3x^2-12x+9$

For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow 6x-12=0$
$\Rightarrow x=2$.

Now,

$m\left(0\right)=9,m\left(2\right)=-3$ and $m\left(5\right)=24$
Hence,

The maximum value is attained at $x=5$.

Hence, option D.