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

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

Given \(f(x)\) is a product, we apply the product rule using \(g(x) = (x - 2)^{2/3}\) and \(h(x) = 2x + 1\).

The product rule states:

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

The derivative of \(g(x)\) is:

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

The derivative of \(h(x)\) is:

\(h'(x) = 2\)

Substituting into the product rule:

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

After simplification:

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

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

The derivative \(f'(x)\) is undefined at \(x = 2\), due to the term \((x - 2)^{-1/3}\).

Next, we solve for \(x\) when \(f'(x) = 0\):

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

By factoring and simplifying, we find that critical points occur at \(x = 1\) (where \(f'(x) = 0\)) and \(x = 2\) (where \(f'(x)\) is undefined).

Conclusion: The function \(f(x) = (x - 2)^{2/3}(2x + 1)\) possesses two critical points: one stemming from \(f'(x) = 0\) and another from \(f'(x)\) being undefined.

Therefore, there are 2 critical points.