## Calculus with Applications (10th Edition)

$$f(x)=(x+3)^{4}$$ Critical number: $-3$. the second derivative test gives no information about extrema, by using the first derivative test, we find that: $-3$ leads to a relative minimum.
$$f(x)=(x+3)^{4}$$ First, find the points where the derivative is $0$ Here \begin{aligned} f^{\prime}(x) &=(4)(1)(x+3)^{3}\\ &=4(x+3)^{3}\\ \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x) =4(x+3)^{3} &=0\\ (x+3) &=0\\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ x =-3 \end{aligned} Critical number: $-3$. Now use the second derivative test. The second derivative is $$f^{\prime\prime}(x)=4(3)(x+3)^{2}=12(x+3)^{2}.$$ Evaluate $f^{\prime\prime}(x)$ at $-3$, getting $$f^{\prime\prime}(-3)=12((-3)+3)^{2}=0 .$$ then the test gives no information about extrema, so we can use the first derivative test. Check the sign of $f^{\prime}(x)$ in the intervals $$(-\infty, -3 ), \quad (-3,\infty).$$ (1) Test a number in the interval $(-\infty, -3 )$ say $-4$ \begin{aligned} f^{\prime}(-4) =4((-4)+3)^{3} \\ &=4(-1)^{3} \\ &=-4 \\ &\lt 0 \end{aligned} to see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, -3 )$ (2) Test a number in the interval $(-3, \infty )$ say $0$ \begin{aligned} f^{\prime}(0) =4((0)+3)^{3} \\ &=4(3)^{3} \\ &=108 \\ &\gt 0 \end{aligned} to see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-3, \infty ).$ so that by the first derivative test, $-3$ leads to a relative minimum.