Extrema and MVT

Given that  $f\left(x\right)$  is a continuous function, if  $f'\left(x\right)<0$  over a certain interval, then  $f\left(x\right)$  is _________ over that interval.

• decreasing

• increasing

• concave up

• concave down

Given that  $f\left(x\right)$  is a continuous function, if  $f'\left(x\right)>0$  over an interval, then  $f\left(x\right)$  is __________ over that interval.

• increasing

• decreasing

• concave up

• concave down

Use the sign chart for f'(x).
There is(are) ...

• a local maximum at x = -2.
• a local maximum at x = 4.
• local maxima at x = -2 and x = 4.
• no extrema.

Which value is guaranteed by the MVT for the function given: f(x) = x² - 6x + 8 on [0,3]

• f(c) = 10

• f(c) = 4

• f ' (c) = -3

• f ' (c) = 7/3

Use the sign chart for f'(x).  There is(are) ...

• a local maximum at x = -2.
• a local minimum at
  x = -2 and a local maximum at x = 4.
• a local maximum at
  x = -2 and a local minimum at x = 4.
• no extrema.

Find all values of c that satisfy the MVT for the function on the given interval

f(x) = -x² + 8x - 17 on [2,6]

• c = 2

• c = 3

• c = 4

• c = 6

Tags

CCSS.HSA.REI.D.11

Find the absolute maximum for the function on the given interval.

f(x) = x³ + 6x² + 9x + 3 on [-4,0]

• -4

• -1

• 0

• 3

 $f\left(x\right)=6x^2-x^3$  

Find where  $f\left(x\right)$   is decreasing.

• (-4,0)

• (0,4)

•  $\left(-\infty,0\right),\left(4,\infty\right)$  

•  $\left(-\infty,-4\right),\left(0,\infty\right)$  

If f '(3) = 0 and f "(3) < 0, then which of the following must be true for f(x)?

• There is a local max at x = 3

• There is a local min at x = 3

• There is an inflection point at x = 3

• There is an x-intercept at x = 3

Find where $f\left(x\right)=6x^2-x^3$  is decreasing.

•  $\left(-\infty,\ 0\right]$  

•  $\left[0,\ 4\right]$  

•  $\left(-\infty,\ 0\right]\ \cup\left[4,\ \infty\right]$  

•  $\left(-\infty,\ -4\right]\cup\left[0,\ \infty\right)$