Question

The total number of local maxima and local minima of the function

$f(x) = \begin{cases} (2+x)^3, & -3 < x \leq -1 \\ x^{2/3}, & -1 < x < 2 \end{cases}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

The given function is a piecewise function:

$f(x) = \begin{cases} (2+x)^3, & -3 < x \leq -1 \\ x^{2/3}, & -1 < x < 2 \end{cases}$

To find the local maxima and minima, we need to analyze the function in each interval and at the point where the definition changes ($x = -1$).

1. Analysis for $-3 < x \leq -1$:

Here, $f(x) = (2+x)^3$.
First derivative: $f'(x) = 3(2+x)^2$.
Set $f'(x) = 0 \implies 3(2+x)^2 = 0 \implies x = -2$.
Let's check the sign of $f'(x)$ around $x = -2$:

• For $x < -2$ (e.g., $x = -2.5$), $f'(x) = 3(2-2.5)^2 = 3(-0.5)^2 = 0.75 > 0$.
• For $x > -2$ (e.g., $x = -1.5$), $f'(x) = 3(2-1.5)^2 = 3(0.5)^2 = 0.75 > 0$.

Since $f'(x)$ does not change sign around $x = -2$, $x = -2$ is an inflection point, not a local extremum.
In fact, $f'(x) \geq 0$ for all $x$ in this interval, meaning $f(x)$ is non-decreasing. It is strictly increasing except at $x=-2$.

2. Analysis for $-1 < x < 2$:

Here, $f(x) = x^{2/3}$.
First derivative: $f'(x) = \frac{2}{3}x^{(2/3)-1} = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}}$.
$f'(x) = 0$ has no solution.
$f'(x)$ is undefined when $x^{1/3} = 0$, which means $x = 0$. This is a critical point within the interval $(-1, 2)$.
Let's check the sign of $f'(x)$ around $x = 0$:

• For $x \in (-1, 0)$ (e.g., $x = -0.5$), $f'(x) = \frac{2}{3(-0.5)^{1/3}}$. Since $(-0.5)^{1/3}$ is negative, $f'(x) < 0$. So, $f(x)$ is decreasing.
• For $x \in (0, 2)$ (e.g., $x = 0.5$), $f'(x) = \frac{2}{3(0.5)^{1/3}}$. Since $(0.5)^{1/3}$ is positive, $f'(x) > 0$. So, $f(x)$ is increasing.

Since $f'(x)$ changes from negative to positive at $x = 0$, and $f(x)$ is continuous at $x=0$, $x = 0$ is a point of local minimum.
The local minimum value is $f(0) = 0^{2/3} = 0$.

3. Analysis at the junction point $x = -1$:

First, check continuity at $x = -1$:

• Left-hand limit: $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (2+x)^3 = (2-1)^3 = 1^3 = 1$.
• Right-hand limit: $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} x^{2/3} = (-1)^{2/3} = ((-1)^2)^{1/3} = 1^{1/3} = 1$.
• Function value: $f(-1) = (2-1)^3 = 1^3 = 1$.

Since the left-hand limit, right-hand limit, and function value are all equal, $f(x)$ is continuous at $x = -1$.

Now, check the behavior of the derivative around $x = -1$:

• For $x < -1$ (in $(-3, -1)$), $f'(x) = 3(2+x)^2$. As $x \to -1^-$, $f'(x) \to 3(2-1)^2 = 3 > 0$. This means $f(x)$ is increasing as it approaches $x = -1$ from the left.
• For $x > -1$ (in $(-1, 2)$), $f'(x) = \frac{2}{3x^{1/3}}$. As $x \to -1^+$, $f'(x) \to \frac{2}{3(-1)^{1/3}} = \frac{2}{3(-1)} = -\frac{2}{3} < 0$. This means $f(x)$ is decreasing as it moves away from $x = -1$ to the right.

Since the function changes from increasing to decreasing at $x = -1$, $x = -1$ is a point of local maximum.
The local maximum value is $f(-1) = 1$.

Conclusion:

• We found a local maximum at $x = -1$.
• We found a local minimum at $x = 0$.

The total number of local maxima and local minima is $1 + 1 = 2$.