Question

1. if the derivative of $f$ is given by $f(x)=e^{x}-3x^{2}$, at which of the following values of $x$ does $f$ have a relative maximum value? (a) $-0.46$ (b) $0.20$ (c) $0.91$ (d) $0.95$ (e) $3.73$

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. \(0.95\)