Question

Find all points of relative extrema, and use a line to demonstrate. y=-x^3-11x^2-40x-49

Answer 1

Step 1

Given;

$y=-x^3-11x^2-40x-49$

Required; To find the relative extrema

Step 2

Use the first derivative test. Find f'(x)

$f^{\prime}(x)=-3x^2-22x-40$

Determine the critical point

To do this we set f'(x)=0

$\begin{gathered} -3x^2-22x-40=0 \\ Using\text{ the quadratic formula} \\ x_{1,\:2}=\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(-3\right)\left(-40\right)}}{2\left(-3\right)} \\ x_{1,\:2}=\frac{-\left(-22\right)\pm \:2}{2\left(-3\right)} \\ x_1=\frac{-\left(-22\right)+2}{2\left(-3\right)},\:x_2=\frac{-\left(-22\right)-2}{2\left(-3\right)} \\ x=-4,\:x=-\frac{10}{3} \\ There\text{ are no domain restrictions, therefore both will be solutions to the critical points} \end{gathered}$

Find the y values

$\begin{gathered} f(-4)=_-(-4)^3-11(-4)^2-40(-4)-49=-1 \\ f(-\frac{10}{3})=-(-\frac{10}{3})^3-11(-\frac{10}{3})^2-40(-\frac{10}{3})-49=-\frac{23}{27} \end{gathered}$

Therefore ;

$\mathrm{Minimum}\left(-4,\:-1\right),\:\mathrm{Maximum}\left(-\frac{10}{3},\:-\frac{23}{27}\right)$