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

Question

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

Accepted Answer

To find the number of critical points of the function $f(x)=(x-2)^{2 / 3}(2 x+1)$, we need to determine where its derivative $f'(x)$ is equal to zero or undefined. Critical points occur where the derivative is zero or does not exist.

First, let's find the derivative of the function:

$f(x)=(x-2)^{2 / 3}(2 x+1)$

We apply the product rule for differentiation, which states that $(uv)' = u'v + uv'$, where $u = (x-2)^{2/3}$ and $v = 2x + 1$.

We need the derivatives of $u$ and $v$:

For $u = (x-2)^{2/3}$, we use the chain rule:

$$u'(x) = \frac{d}{dx}[(x-2)^{2/3}] = \frac{2}{3}(x-2)^{-1/3} \cdot 1 = \frac{2}{3}(x-2)^{-1/3}$$

The derivative of $v$ is straightforward, as $v = 2x + 1$:

$v'(x) = 2$

Now we apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substituting $u$, $u'$, and $v$, we get:

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

This simplifies to:

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

For critical points, we need to solve $f'(x) = 0$ or where it is undefined.

1. Solve for where the derivative is zero:

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

Combining like terms in a common denominator, we get:

$$\frac{2(2x + 1) + 6(x - 2)}{3(x-2)^{1/3}} = 0$$

Simplifying the numerator:

$$2(2x + 1) + 6(x-2) = 4x + 2 + 6x - 12 = 10x - 10 = 10(x-1)$$

So:

$$\frac{10(x-1)}{3(x-2)^{1/3}} = 0$$

The numerator is zero when:

$10(x-1) = 0$

Therefore:

$x = 1$

2. Solve for where the derivative is undefined:

The denominator, $3(x-2)^{1/3}$, is undefined when $(x-2)^{1/3} = 0$, which happens at:

$x = 2$

From the above analysis, the critical points are at $x = 1$ and $x = 2$. Thus, there are 2 critical points.

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

Option A

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