Find the Critical Points f(x)=(x^2-4)^(2/3)

Problem

ƒ(x)=(x^2−4)^(2/3)

Solution

1. Identify the definition of critical points, which occur where the first derivative ƒ(x)^′ is equal to zero or where the derivative is undefined.

2. Apply the chain rule to find the derivative of the function ƒ(x)=(x^2−4)^(2/3)

ƒ(x)^′=2/3*(x^2−4)^(−1/3)⋅d(x^2−4)/d(x)

3. Differentiate the inner function x^2−4 and simplify the expression.

ƒ(x)^′=2/3*(x^2−4)^(−1/3)⋅2*x

ƒ(x)^′=(4*x)/(3*(x^2−4)^(1/3))

4. Solve for x where ƒ(x)^′=0 by setting the numerator equal to zero.

4*x=0

x=0

5. Determine where ƒ(x)^′ is undefined by setting the denominator equal to zero.

3*(x^2−4)^(1/3)=0

x^2−4=0

x^2=4

x=2,x=−2

6. Verify that these x values are in the domain of the original function ƒ(x) Since ƒ(x) is defined for all real numbers, all three values are critical points.

Final Answer

x=−2,0,2

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