The function $f(x)=2 x^3-15 x^2+36 x+4$ is maximum at

(A) $x=2$

(B) $x=4$

(C) $x=0$

(D) $x=3$

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Best answer

**SOLUTION —**

$\begin{array}{l}f(x)=2 x^3-15 x^2+36 x+4 \\f^{\prime}(x)=6 x^2-30 x+36\end{array}$

Either maxima or minima $f^{\prime}(x)=0$

$\begin{array}{l}\therefore \quad 6 x^2-30 x+36=0 \\\Rightarrow \quad 6(x-2)(x-3)=0 \Rightarrow x=2,3 \\\end{array}$

Now,

$\begin{aligned}\Rightarrow f^{\prime \prime}(x) & =12 x-30 \\f^{\prime \prime}(2)=24-30 & =-6<0\end{aligned}$

Therefore, $f(x)$ is maximum at, $x=2$

So, The correct option will be **(A).**

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