Question

The number of critical points of the function $f(x) = (x - 2)^{2/3}(2x + 1)$ is:

• A. 2
• B. 0
• C. 1
• D. 3

Answer 1

The Correct Option is A

To find the number of critical points of the function \(f(x) = (x - 2)^{2/3}(2x + 1)\), we need to determine the points where the derivative \(f'(x)\) is either zero or undefined.

Step 1: Find the derivative \(f'(x)\).

The product rule is applicable here since \(f(x)\) is a product of two functions: \(g(x) = (x - 2)^{2/3}\) and \(h(x) = (2x + 1)\).

The derivative is given by:

\(f'(x) = g'(x)h(x) + g(x)h'(x)\)

First, find \(g'(x)\) where \(g(x) = (x - 2)^{2/3}\):

\(g'(x) = \frac{2}{3}(x - 2)^{-1/3}\)

Next, find \(h'(x)\) where \(h(x) = 2x + 1\):

\(h'(x) = 2\)

Substituting these into the product rule formula:

\({f'(x) = \frac{2}{3}(x - 2)^{-1/3}(2x + 1) + (x - 2)^{2/3} \cdot 2}\)

Simplifying the expression, we have:

\(f'(x) = \frac{2(2x + 1)}{3(x - 2)^{1/3}} + 2(x - 2)^{2/3}\)

Step 2: Determine where \(f'(x) = 0\) or where \(f'(x)\) is undefined.

The expression is undefined at \(x = 2\) because \((x - 2)^{-1/3}\) becomes undefined.

Next, set \(f'(x) = 0\) and solve for \(x\):

\(\frac{2(2x + 1)}{3(x - 2)^{1/3}} + 2(x - 2)^{2/3} = 0\)

Factor out common terms and simplify to find the possible roots (you may need to equate numerators if possible).

After simplification, the critical points arise at \(x = 1\) (from solving \(f'(x) = 0\)) and \(x = 2\) (as \(f'(x)\) is undefined).

Conclusion: The function \(f(x) = (x - 2)^{2/3}(2x + 1)\) has two critical points: one where the derivative is zero and one where it is undefined.

Thus, the number of critical points is 2.

Answer 2

Solution:

The given function is \(f(x)\). Its derivative is:

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

Simplify the numerator:

\[ f'(x) = \frac{2}{3} \cdot \frac{(2x + 1) + 3(x - 2)}{(x - 2)^{1/3}}. \]

Expand and simplify:

\[ (2x + 1) + 3(x - 2) = 5x - 5. \]

Thus:

\[ f'(x) = \frac{2(5x - 5)}{3(x - 2)^{1/3}}. \]

Critical points:

1. Set \(f'(x) = 0\):
2. The derivative \(f'(x)\) is undefined at \(x = 2\).

Hence, the critical points are:

\[ x = 1 \quad \text{and} \quad x = 2. \]

Final Answer: 2.