GATE DS&AI 2024 | Question: 40

Question

​​​​​Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ where $\mathbb{R}$ is the set of all real numbers.

\[
f(x)=\frac{x^{4}}{4}-\frac{2 x^{3}}{3}-\frac{3 x^{2}}{2}+1
\]

Which of the following statements is/are TRUE?

• A. $x=0$ is a local maximum of $f$
• B. $x=3$ is a local minimum of $f$
• C. $x=-1$ is a local maximum of $f$
• D. $x=0$ is a local minimum of $f$

Answer 1

Answer:

x=0 is local maximum

x=3 is local minimum

Answer 2

Answer: A,B

Given function:

\[ f(x) = \frac{x^4}{4} - \frac{2x^3}{3} - \frac{3x^2}{2} + 1 \]

First, we'll find the critical points by finding where the derivative is zero:

\[ f'(x) = x^3 - 2x^2 - 3x   = x(x^2 - 2x - 3) = x(x - 3)(x + 1) \]

So, the critical points are \( x = -1, 0, 3 \).

\[ f''(x) = 3x^2 - 4x - 3 \]

Now, we need to find the behavior around these points:

1. At \( x = -1 \):

\[ f''(-1) = 3(-1)^2 - 4(-1) - 3 \]

\[ f''(-1) = 3(1) + 4 - 3 \]

\[ f''(-1) = 4 > 0 \]

2. At \( x = 0 \):

\[ f''(0) = 3(0)^2 - 4(0) - 3 \]

\[ f''(0) = -3 < 0 \]

3. At \( x = 3 \):

\[ f''(3) = 3(3)^2 - 4(3) - 3 \]

\[ f''(3) = 27 - 12 - 3 \]

\[ f''(3) = 12 > 0 \]

$x=-1$ is a local minimum of $f$

$x=0$ is a local maximum of $f$

$x=3$ is a local minimum of $f$