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

MCQ

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

A
$3$
B
$0$
C
$4$
✓
$2$

✓

Answer

Correct option: D.

$2$

Given, $f(x)=2 x^3-15 x^2+36 x+4$
Implies that $f^{\prime}(x)=6 x^2-30 x+36$
For a local maxima or a local minima, we must have $f^{\prime}(x)=0$
Implies that $6 x^2-30 x+36=0$
Implies that $x^2-5 x+6=0$
$(x-2)(x-3)=0$
Implies that $x=2,3$
Now, $f^{\prime \prime}(x)=12 x-30$
Implies that $f^{\prime \prime}(2)=24-30=6 < 0$
Therefore, $x=1$ is a local maxima.
Also, $f^{\prime \prime}(3)=36-30=6 > 0$
Therefore, $x=2$ is a local maxima.

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