Consider the function f(x)=2x^3−15x^2−36x+8 on the interval [−6,8].

The average or mean slope of the function on this interval is equal to : 38

By the Mean Value Theorem, we know there exists a c in the open interval (−6,8) such that f′(c) is equal to this mean slope. For this problem, there are two values of c that work:

The smaller one c1= ?

The larger one c2= ?

Can someone help me finish this problem?

Question

Consider the function f(x)=2x^3−15x^2−36x+8 on the interval [−6,8].

The average or mean slope of the function on this interval is equal to  : 38

By the Mean Value Theorem, we know there exists a c in the open interval (−6,8) such that f′(c) is equal to this mean slope. For this problem, there are two values of c that work:

The smaller one c1= ?

The larger one c2= ?

Can someone help me finish this problem?

lalaly · Answer

find the derivative of f(x) and set it equal to the mean slope and solve for x
the x's u get are c1 and c2

anonymous · Answer

I got 6(x^2-5x-6)=38 then I'm stuck at (x-6)(x+1)=32 I don't know how to solve this...

lalaly · Answer

no $$6x^2-30x-36=38$$$$6x^2-30x-74=0$$

lalaly · Answer

now use the quadratic formula

anonymous · Answer

I get a negative number inside of sqrt thingy...