Question

Consider the function f(x) = 2 x^3 - 3 x^2 - 72 x + 7 on the interval [ -4 , 8 ]. Find the average or mean slope of the function on this interval. _____

By the Mean Value Theorem, we know there exists a c in the open interval ( -4 , 8 ) such that f'( c) is equal to this mean slope. For this problem, there are two values of c that work. The smaller one is ______and the larger one is ________

Answer 1

Have you considered calculating f(-4) and f(8)?

Answer 2

yes i did and i got the mean slope which is 12 but i cant get the other 2 values

Answer 3

i got -3.27 and 4.27 as the answers but there wrong :(

Answer 4

i took the first derivative and set it equal to 12

Answer 5

Whoops! Why did I write f(x)?! Of course you are right.

Let's see...

f'(x) = 6x^2 - 6x - 72 x

6x^2 - 6x - 72 = 12

x^2 - x - 12 = 2

x^2 - x - 14 = 0

x = \((1/2)(\sqrt{57}+1)\;or\;(1/2)(1-\sqrt{57})\)

That looks like 4.275... or -3.275... to me. More decimal places, maybe? Exact values maybe? You totally have it. We're just trying to figure out why your system doubts you.

Answer 6

hmmm not too sure then:S it is an online thing that wont let me write exact values

Answer 7

Best I can do, then, is wish you good luck. Maybe in a different order?

Answer 8

will do thanks for the help:)