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

Problem

ƒ(x)=(x2−4)(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)=(x2−4)(2/3)

ƒ(x)′=2/3*(x2−4)(−1/3)⋅d(x2−4)/d(x)

3. Differentiate the inner function x2−4 and simplify the expression.

ƒ(x)′=2/3*(x2−4)(−1/3)⋅2*x

ƒ(x)′=(4*x)/(3*(x2−4)(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*(x2−4)(1/3)=0

x2−4=0

x2=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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