#### Question

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

(a) 3

(b) 0

(c) 4

(d) 2

#### Solution

(d) 2

\[\text { Given }: f\left( x \right) = 2 x^3 - 15 x^2 + 36x + 4\]

\[ \Rightarrow f'\left( x \right) = 6 x^2 - 30x + 36\]

\[\text { For a local maxima or a local minima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 6 x^2 - 30x + 36 = 0\]

\[ \Rightarrow x^2 - 5x + 6 = 0\]

\[ \Rightarrow \left( x - 2 \right)\left( x - 3 \right) = 0\]

\[ \Rightarrow x = 2, 3\]

\[\text { Now }, \]

\[f''\left( x \right) = 12x - 30\]

\[ \Rightarrow f''\left( 2 \right) = 24 - 30 = - 6 < 0\]

\[\text { So, x = 1 is a local maxima } . \]

\[\text { Also }, \]

\[f''\left( 3 \right) = 36 - 30 = 6 > 0\]

\[\text { So,x=2 is a local maxima }.\]

Is there an error in this question or solution?

Solution The Function F(X) = 2 X 3 − 15 X 2 + 36 X + 4 is Maximum at X = (A) 3 (B) 0 (C) 4 (D) 2 Concept: Graph of Maxima and Minima.