Question

determine whether rolles theorem applies to the given function on the given interval. if so, find the point(s) that are guaranteed to exist by rolles theorem.

$f(x)=(x - 1)^4-9(x - 1)^2+8;-2,1$

since f is continuous on the closed interval $-2,1$, differentiable on the open interval $(-2,1)$, and $f(-2)=f(1)$, rolles theorem can be applied to f on the interval $-2,1$.

select the correct choice and, if necessary, fill in the answer box to complete your choice.

a. rolles theorem applies, and the point(s) it guarantees to exist is/are $c=-1.12$
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)
b. rolles theorem cannot be applied.

Explanation:

Step1: Find the derivative of \(f(x)\)

Let \(u = x - 1\), then \(f(x)=u^{4}-9u^{2}+8\). Using the chain - rule \(\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}\).
First, \(\frac{df}{du}=4u^{3}-18u\) and \(\frac{du}{dx} = 1\). So \(f^{\prime}(x)=4(x - 1)^{3}-18(x - 1)=2(x - 1)[2(x - 1)^{2}-9]=2(x - 1)(2x^{2}-4x+2 - 9)=2(x - 1)(2x^{2}-4x - 7)\)

Step2: Set \(f^{\prime}(c)=0\)

Since Rolle's Theorem applies ( \(f(x)\) is continuous on \([-2,1]\), differentiable on \((-2,1)\) and \(f(-2)=f(1)\)), we set \(f^{\prime}(c)=0\).
We have two cases from \(2(c - 1)(2c^{2}-4c - 7)=0\):

• Case 1: \(c - 1=0\), but \(c = 1

otin(-2,1)\)

• Case 2: \(2c^{2}-4c - 7=0\). Using the quadratic formula \(c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 2\), \(b=-4\), \(c=-7\))

$$c=\frac{4\pm\sqrt{16+56}}{4}=\frac{4\pm\sqrt{72}}{4}=\frac{4\pm6\sqrt{2}}{4}=1\pm\frac{3\sqrt{2}}{2}$$

Since \(c\in(-2,1)\), \(c=1-\frac{3\sqrt{2}}{2}\approx1 - 2.12=-1.12\)

Answer:

A. Rolle's Theorem applies, and the point(s) it guarantees to exist is/are \(c = 1-\frac{3\sqrt{2}}{2}\)