How Does the Mean Value Theorem Apply to Finding the Slope on an Interval?

[Wm_Davies]

Homework Statement

Consider the function f(x)=2x^3−12x^2−72x+6 on the interval [−4,7] . Find the average or mean slope of the function on this interval.

Homework Equations

MEAN VALUE THEOREM
f'(c) = \frac{f(b)-f(a)}{b-a}

The Attempt at a Solution

When I set this problem up in the mean value theorem I found that

6c^2 - 24c - 72 = (-400+26)/(11)
Finally giving me

6c^2 - 24c -38 = 0

Solving with the quadriatic equation I got

2\pm\frac{\sqrt93}{3}


Which I know is correct, but the program says it is not correct so maybe the question is not asking for me to do this...?

 

[n!kofeyn]

I am honestly not for sure what you are doing. You just have some equations there. What the mean value theorem says is that on a closed interval [a,b], there exists a point c such that f'(c)="mean slope of f over the interval [a,b]". What equation gives you the mean slope?

 

[Wm_Davies]

I am honestly not for sure what you are doing. You just have some equations there. What the mean value theorem says is that on a closed interval [a,b], there exists a point c such that f'(c)="mean slope of f over the interval [a,b]". What equation gives you the mean slope?

I was finding the value of "c" in the mean value theorem, because I was overthinking this problem. After I read your post several times it finally sunk in that the mean slope is equal to (y1-y0)/(x1-x0). Which is -34 and the answer to the problem. Thank you for the help.

 

[n!kofeyn]

Great! Yes, (f(b)-f(a))/(b-a) is the mean slope over the interval [a,b]. Just to reiterate, the mean value theorem guarantees there is at least one point c such that f'(c) is exactly equal to the mean slope over [a,b].