4. if the derivative of $f$ is given by $f(x)=e^{x}-3x^{2}$, at which of the following values of $x$ does…

Explanation:

Step1: Recall the first - derivative test

A function (y = f(x)) has a relative maximum at a point (x = c) if (f^{\prime}(c)=0) and (f^{\prime}(x)) changes sign from positive to negative at (x = c).

Step2: Analyze the sign of (f^{\prime}(x)=e^{x}-3x^{2}) for each option

• For (x=-0.46): (f^{\prime}(-0.46)=e^{- 0.46}-3(-0.46)^{2}) (=\frac{1}{e^{0.46}}-3\times0.2116) (\approx\frac{1}{1.5849}-0.6348) (\approx0.631 - 0.6348<0)

• For (x = 0.20): (f^{\prime}(0.20)=e^{0.20}-3(0.20)^{2}) (\approx1.2214 - 3\times0.04) (=1.2214 - 0.12>0)

• For (x = 0.91): (f^{\prime}(0.91)=e^{0.91}-3(0.91)^{2}) (\approx2.488 - 3\times0.8281) (=2.488-2.4843\approx0.0037>0)

• For (x = 0.95): (f^{\prime}(0.95)=e^{0.95}-3(0.95)^{2}) (\approx2.586 - 3\times0.9025) (=2.586 - 2.7075<0)

• For (x = 3.73): (f^{\prime}(3.73)=e^{3.73}-3(3.73)^{2}) (\approx41.7 - 3\times13.9129) (=41.7-41.7387<0)

We check the sign change of (f^{\prime}(x)). We know that (f^{\prime}(x)) is continuous. We can also use the fact that if (f^{\prime}(x)) changes from positive to negative at (x = c), then (f(x)) has a relative maximum at (x = c). We check the intervals: Let's consider the interval around (x = 0.95). Take a value slightly less than (0.95), say (x = 0.94) (f^{\prime}(0.94)=e^{0.94}-3(0.94)^{2}\approx2.56 - 3\times0.8836=2.56 - 2.6508=- 0.0908) Take a value slightly greater than (0.95), say (x = 0.96) (f^{\prime}(0.96)=e^{0.96}-3(0.96)^{2}\approx2.61 - 3\times0.9216=2.61 - 2.7648=-0.1548)

Wait, no. We should check the sign change from positive to negative. We know that (f^{\prime}(0.91)>0) and (f^{\prime}(0.95)<0)

Answer:

D. (0.95)