The function

Question:

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

(a) 3

(b) 0

(c) 4

(d) 2

Solution:

(d) 2

Given : $f(x)=2 x^{3}-15 x^{2}+36 x+4$

$\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 6 x^{2}-30 x+36=0$

$\Rightarrow x^{2}-5 x+6=0$

$\Rightarrow(x-2)(x-3)=0$

$\Rightarrow x=2,3$

Now,

$f^{\prime \prime}(x)=12 x-30$

$\Rightarrow f^{\prime \prime}(2)=24-30=-6<0$

So, $x=1$ is a local maxima.

Also,

$f^{\prime \prime}(3)=36-30=6>0$

So, $x=2$ is a local maxima.

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