5. (5 points) which of the following functions has a critical value at ( x = 0 )? be careful to consider the…

Explanation:

Step1: Recall the definition of critical value

A critical value of a function (y = f(x)) is a value of (x) in the domain of (f) where (f^{\prime}(x)=0) or (f^{\prime}(x)) does not exist.

Step2: Differentiate function (A)

Differentiate (f(x)=x^{3}-3x^{2}+2) using the power rule ((x^{n})^{\prime}=nx^{n - 1}). Then (f^{\prime}(x)=3x^{2}-6x). Set (f^{\prime}(x) = 0), so (3x^{2}-6x=3x(x - 2)=0). Solving (3x(x - 2)=0) gives (x = 0) or (x=2). Since (x = 0) is in the domain ((-1,\infty)), (x = 0) is a critical value for (f(x)).

Step3: Check other functions for completeness

• Differentiate (g(x)=x^{4}-4x^{3}+6x) using the power rule. (g^{\prime}(x)=4x^{3}-12x^{2}+6). Then (g^{\prime}(0)=4\times0^{3}-12\times0^{2}+6=6\neq0).
• Differentiate (h(x)=\ln(x + 10)) using the formula ((\ln(u))^{\prime}=\frac{u^{\prime}}{u}), where (u=x + 10) and (u^{\prime}=1). So (h^{\prime}(x)=\frac{1}{x + 10}). Then (h^{\prime}(0)=\frac{1}{0+10}=\frac{1}{10}\neq0).
• Differentiate (k(x)=e^{x}-x^{2}) using the rules ((e^{x})^{\prime}=e^{x}) and ((x^{n})^{\prime}=nx^{n-1}). So (k^{\prime}(x)=e^{x}-2x). Then (k^{\prime}(0)=e^{0}-2\times0=1\neq0).

Answer:

A. (f(x)=x^{3}-3x^{2}+2)