Find the relative extrema, if any, of $f(t)=e^{t}-7 t-3$ . Use the Second Derivative Test, if possible.
A) relative minimum: none, relative maximum: $f(\ln 3)=-7 \ln 7$
B) relative minimum: $f(\ln 7)=4-7 \ln 7$ , relative maximum: none
C) relative minimum: $f(\ln 3)=-7 \ln 7$ , relative maximum: none
D) relative minimum: none, relative maximum: $f(\ln 7)=4-7 \ln 7$

Question

Quizplus · Accepted Answer

To find relative extrema, first find the derivative: $f'(t) = e^t - 7$. Set $f'(t) = 0$ to find critical points: $e^t = 7$, so $t = \ln 7$. Use the second derivative: $f''(t) = e^t$. Since $f''(\ln 7) = 7 > 0$, there is a relative minimum at $t = \ln 7$. Evaluate $f(\ln 7)$ to find the minimum value. There is no relative maximum.

Find the Relative Extrema, If Any, Of f(t)=et−7t−3f(t)=e^{t}-7 t-3f(t)=et−7t−3 Use the Second Derivative Test, If Possible

Find the Relative Extrema, If Any, Of $f(t)=e^{t}-7 t-3$ Use the Second Derivative Test, If Possible