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)=(-\left(-22\right)\pm √(\left(-22\right)^2-4\left(-3\right)\left(-40\right)))/(2\left(-3\right)) \\ x_(1,\:2)=(-\left(-22\right)\pm \:2)/(2\left(-3\right)) \\ x_1=(-\left(-22\right)+2)/(2\left(-3\right)),\:x_2=(-\left(-22\right)-2)/(2\left(-3\right)) \\ x=-4,\:x=-(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(-(10)/(3))=-(-(10)/(3))^3-11(-(10)/(3))^2-40(-(10)/(3))-49=-(23)/(27) \end{gathered}`

Therefore ;

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