f(x) = -3x^3 - 1x^2 + 4x + 4?

Question

Find the average slope of this function on the interval ( -2 , -1 ).

By the Mean Value Theorem, we know there exists a c in the open interval ( -2, -1 ) such that f ' (c) is equal to this mean slope. Find the value of c in the interval which works

Answer 1

f'(x) = -9x^2-2x+4

f(-2) = 16

f(-1)= 2

(f(-1) - f(-2))/(-1-(-2)) = -14

Set f'(x) = -14 = -9x^2-2x+4

0 = -9x^2-2x+18 = x^2+2/9x-2 --> x = -1.53

Answer 2

The slope is found by f'(x). Since you are looking for the average slope, then

slope_ave = 1/(b-a)*integral(f'(x),a,b) = 1/(b-a)*(f(b)-f(a))
now a = -2 and b = -1
slope_ave = 1/(-1 - -2)*(2 - 16) = -14

then find f'(c) = -14

f'(c) = -9c^2 - 2c + 4 = -14
or
-9c^2 - 2c + 18 = 0
then c = (2 +- sqrt(4 - 4*(-9)*18))/(-18)
c = (2 +- 25.5343)/(-18)
so c = -1.5297 or 1.3075
since c in [a,b], then c = -1.5297

Answer 3

can you rewrite the question using * for 'times'? I don't know if 3x3 is three times three or 3 X 3, where X is unknown value. also, the second line: are these two different expressions or what is with the space between 8 and 6????