Mean-Value Theorem Problem

[turophile]

The problem:

Let g(x) = x[sup:10vuzol6]4[/sup:10vuzol6] – 4x[sup:10vuzol6]3[/sup:10vuzol6] – 2x[sup:10vuzol6]2[/sup:10vuzol6] + 12x + 7. Show that the equation g'(x) = – 8 has a solution in the interval (1, 3).

My work so far:

Consider the function h(x) = g(x) + 8x. Then h(x) is also continuous on [1, 3] and differentiable on its interior. By the MVT, there is a point q in the interior such that h(3) – h(1) = (3 – 1)h'(q) = 2h'(q).

h'(x) = g'(x) + d/dx (8x) = 4x[sup:10vuzol6]3[/sup:10vuzol6] – 12x[sup:10vuzol6]2[/sup:10vuzol6] – 4x + 12 + 8 = 4(x[sup:10vuzol6]3[/sup:10vuzol6] – 3x[sup:10vuzol6]2[/sup:10vuzol6] – x + 3) + 8
= 4(x + 1)(x – 1)(x – 3) + 8 ? – 8 + 8 = 0

? x = – 1, 1, or 3

I think I'm on the right track here, but I'm not sure what the next step is. Any hints?

 

[mmm4444bot]

I'll try to remember this one.

We need to show that the slope of function g is -8 at some point between x = 1 and x = 3.

To make use of the MVT, we first evaluate the following expression (see Glenn's post below).

[g(3) - g(1)]/[3 - 1] = -8

Therefore, the MVT states that the slope of function g must be -8 at some point between x = 1 and x = 3. And that's exactly what we needed to show.

That was trivial, after correcting my math error. Thank you, Glenn.

 

[BigGlenntheHeavy]

\(\displaystyle g(x) \ = \ x^4-4x^3-2x^2+12x+7, \ g'(x) \ = \ 4x^3-12x^2-4x+12\)

\(\displaystyle MVT: \ g'(c) \ = \ \frac{f(b)-f(a)}{b-a} \ = \ \frac{-2-14}{3-1} \ = \ -8\)

\(\displaystyle Hence \ g'(x) \ = \ -8 \ = \ 4x^{3}-12x^{2}-4x+12\)

\(\displaystyle Therefore \ x \ \dot= \ 2.675\)

\(\displaystyle See \ graph.\)

[attachment=0:1xa6vh45]jjj.jpg[/attachment:1xa6vh45]