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Chapter 0: Problem 52

Determine whether the function is even, odd, or neither. $$ f(x)=2 x^{1 / 3}-3 x^{2} $$

Short Answer

Expert verified
The function \(f(x) = 2x^{1/3} - 3x^2\) is neither even nor odd.

Step by step solution

01

Evaluate \(f(-x)\)

To determine whether the function is even, odd, or neither, we need to evaluate the function at \(-x\) to see if it satisfies any of the conditions. Replace \(x\) with \(-x\) in the given function: \(f(-x) = 2(-x)^{1/3} - 3(-x)^2\)
02

Simplify \(f(-x)\)

Now, simplify the expression obtained in step 1: \(f(-x) = 2(-x)^{1/3} - 3x^2\) Since \((-x)^{1/3} = -x^{1/3}\), the expression becomes: \(f(-x) = -2x^{1/3} - 3x^2\)
03

Check if the function is even

To determine if the function is even, compare \(f(-x)\) with \(f(x)\). \(f(x) = 2x^{1/3} - 3x^2\) \(f(-x) = -2x^{1/3} - 3x^2\) We can conclude that \(f(-x) \neq f(x)\). Since this property does not hold true, the function is not even.
04

Check if the function is odd

To determine if the function is odd, compare \(f(-x)\) with \(-f(x)\). \(f(x) = 2x^{1/3} - 3x^2\) \(-f(x) = -2x^{1/3} + 3x^2\) \(f(-x) = -2x^{1/3} - 3x^2\) We can conclude that \(f(-x) \neq -f(x)\). Since this property does not hold true, the function is not odd.
05

Conclusion

Since the function does not satisfy the properties of being even or odd, it can be concluded that the function \(f(x) = 2x^{1/3} - 3x^2\) is neither even nor odd.

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Most popular questions from this chapter

Write the expression in algebraic form. $$ \cos \left(\sin ^{-1} x\right) $$

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Determine whether \(h=g \circ f\) is even, odd, or neither, given that a. both \(g\) and \(f\) are even. b. \(g\) is even and \(f\) is odd. c. \(g\) is odd and \(f\) is even. d. both \(g\) and \(f\) are odd.

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