GATE Overflow Test Series | Mock GATE | Test 4 | Question: 21

Question

The minimum value of the function $$f(x) = \frac{x^4}{4} - x^2 -3$$ occurs at

• A. $x = 1$
• B. $x =\sqrt 2$
• C. $x = 0$
• D. $ x = \frac{1}{\sqrt{4}}$

Answer 1

$$f(x) = \frac{x^4}{4} - x^2 - 3$$
We will find the first and second derivatives.

$f'(x) = x^3 - 2x$

and $f"(x) = 3x^2 - 2.$

To determine the critical (either maximum or minimum) value of $x,$ equate $f'(x) = 0, \implies  x^3 - 2x = 0 \implies x = 0, \pm \sqrt 2.$

For $x=0, f''(x)$ is negative making it a maximum value. For, $x=\sqrt 2,f''(x)= 6-2=4 >0,$ making $\sqrt 2$ a minimum value.