first derivative test problem

 $f\left(x\right)\ =\ \frac{x^3}{3}+\frac{x^2}{2}-6x+1$  Where is this function increasing or decreasing? Step 1 is find f'  what is it?

•  $x^3-6x^2-9x+1$  

• 9

•  $x^2-2x$  

•  $x^2+x-6$  

The next step is to find the critical values, we set the derivative

 $f'\left(x\right)\ =\ x^2+x-6$  equal to 0 and then find those x values.  What are the critical values? Check tall that apply

• -3

• 0

• 1

• -2

• 2

Tags

CCSS.8.F.A.3

The critical values were -3 and 2, to figure out where the function is rising or falling we need to plug in numbers to the left, in-between and to the right of our critical values. We want to pick "easy numbers" , which are good choices, select all that apply.

• -10

• -13

• 0

• 2

• 10

for values of x < 3, is the derivative positive or negative or zero?

• positive

• negative

• 0

between x = -3 and x = 2 the derivative is ....

• positive

• negative

• 0

we found the derivative just before x= -3 is positive and just after x = -3 is negative, therefore x = -3 is a

• local maximum

• local minimum

• nothing

• absolute maximum

• absolute minimum

just before x = 2 the derivative is negative and just after x = 2 the derivative is positive, therefore x = 2 is a ...

• local maximum

• local minimum

• nothing

• absolute maximum

• absolute minimum