The absolute minimum value of the function f(x) = 4x - 1/2 x^2 in the interval [-2, 9/2] is

Question

The absolute minimum value of the function \(f(x) = 4x - \frac 12x^2\) in the interval \(\left[-2, \frac 92\right]\) is:

(A) −8

(B) −9

(C) −10

(D) −16

Answer 1

Correct option is (C) −10

\(f(x) = 4x - \frac 12 x^2\)

Being a polynomial function f(x) is differentiable \(\forall x \in \left(-2, \frac 92\right)\)

\(f'(x) = 4 -x\)

\(f'(x) = 4 -x = 0\)

\(\Rightarrow x = 4\)

For the function \(f(x) = 4x - \frac 12 x^2\) in the interval \(\left[-2, \frac 92\right]\), the end points are

\(x = -2\ \&\ x = \frac92\)

∴ The absolute minimum value of the function \(f(x) = 4x - \frac 12 x^2\) in the interval \(\left[-2, \frac 92\right]\) is

\(\text{Min}\left \{f(-2), f(4), f(\frac 92)\right\} = \text{Min} \left\{-10, 8, \frac{63}8 \right\} = -10\)