Q. 56 - Calculus

Question

Determine whether or not each function f in Exercises 53–60 satisfies the hypotheses of the Mean Value Theorem on the given interval [a, b]. For those that do, use derivatives and algebra to find the exact values of all c ∈ (a, b) that satisfy the conclusion of the Mean Value Theorem.

$f\left(x\right)=\left(x^2-1\right)\left(x^2-4\right),[a,b]=[-3,3]$

Step-by-Step Solution

Answer

The function satisfies the mean value theorem and $c=0\text{ or }c=\pm \sqrt{2.5}$

Step 1. Given information

$f\left(x\right)=\left(x^2-1\right)\left(x^2-4\right)$

Step 2. Proving Mean Value Theorem

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(c\right)=\frac{f\left(3\right)-f(-3)}{3-(-3)}\\ =\frac{\left(3^2-1\right)\left(3^2-4\right)-\left((-3{)}^2-1\right)\left((-3{)}^2-4\right)}{6}\\ =\frac{8\cdot 5-8\cdot 5}{6}\\ =0\end{array}$

Now,

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{d}{dx}\left(x^2-1\right)\left(x^2-4\right)\\ =\frac{d}{dx}\left(x^4-4x^2-x^2+4\right)\\ =\frac{d}{dx}\left(x^4-5x^2+4\right)\\ =4x^3-10x\end{array}$

Therefore,

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(c\right)=0\\ 4c^3-10c=0\\ 2c\left(2c^2-5\right)=0\\ 2c=0\text{ or }2c^2-5=0\\ c-0\text{ or }{c}^2=\frac{5}{2}\\ c=0\text{ or }c=\pm \sqrt{2.5}\end{array}$