If on [0, 8], find the value c that satisfies the Mean-Value Theorem. (Round to three decimal places.)
A) 1.333
B) 1.540
C) 0.759
D) 1.923
E) 0.385

Question

If  on [0, 8], find the value c that satisfies the Mean-Value Theorem. (Round to three decimal places.)
A) 1.333 
B) 1.540 
C) 0.759 
D) 1.923 
E) 0.385

Quizplus · Accepted Answer

By the Mean-Value Theorem, we know there exists a value $c$ in $(0,8)$ such that $$\frac{f(8)-f(0)}{8-0}=f'(c)$$
Notice that $f(0)=11ead0bc_8ce6_b3c0_99a0_f911d6efbd9e_TB6988_11$ and $f(8)=11ead0bc_8ce6_b3c0_99a0_f911d6efbd9e_TB6988_11+8\cdot 11=11ead0bc_8ce6_b3c0_99a0_f911d6efbd9e_TB6999_11$. Therefore, we have $$\frac{f(8)-f(0)}{8-0}=\frac{11ead0bc_8ce6_b3c0_99a0_f911d6efbd9e_TB6999_11-11ead0bc_8ce6_b3c0_99a0_f911d6efbd9e_TB6988_11}{8}=11$$
We now need to find $c$ such that $f'(c)=11$. To do this, we take the derivative of $f$ and set it equal to 11: $$f'(x)=44x^3-59.040x^2+27.936x+11=11$$
Dividing both sides by 11 gives $$4x^3-5.404x^2+2.540x+1=0$$
Using synthetic division, we find that this polynomial factors as $$(x-1.54)(4x^2+1.2x+0.649)=0$$
Since $0 < c < 8$, the only valid value for $c$ is $c=1.540$, rounded to three decimal places. Thus, the answer is B.

If on [0, 8], Find the Value C That