Question

Find the values of $c$ that satisfy the MVT for integrals on $[2,5]$.
$f\left(z\right)=4z^3-8z^2+7z-2$

• A. $c=\frac{2\pm \sqrt{79}}{3}$
• B. $c=\frac{-2\pm \sqrt{79}}{3}$
• C. $c=\frac{3\pm \sqrt{79}}{3}$
• D. $c=\frac{-3\pm \sqrt{79}}{3}$

Answer 1

The correct option is A $c=\frac{2\pm \sqrt{79}}{3}$

$f\left(z\right)=4z^3-8z^2+7z-2$

$f^{\prime }\left(c\right)=12c^2-16c+7$

$f^{\prime }\left(c\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}$

$=4\left(5^3\right)-8\left(5^2\right)+7\left(5\right)-2-[4\left(2^3\right)-8\left(2^2\right)+7(2)-2]$

$=\frac{500-200+35-2-32+208-14+2}{3}$

$=\frac{497}{3}$

$12c^2-16c+7=\frac{497}{3}$

$12c^2-16c-\frac{476}{3}=0$

$36c^2-48c-476=0$

$c=\frac{48\pm \sqrt{{\left(48\right)}^2-4(-476)\left(36\right)}}{72}$

$c=\frac{48\pm \sqrt{2304+68544}}{72}$

$c=\frac{48\pm 24\sqrt{79}}{72}$

$c=\frac{2\pm \sqrt{79}}{3}$