determine whether rolles theorem applies to the given function on the given interval. if so, find the…

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.\n\n$f(x)=(x - 1)^4-9(x - 1)^2+8;-2,1$\n\nsince $f$ is not continuous on the closed interval $-2,1$, not differentiable on the open interval $(-2,1)$, and $f(-2)=f(1)$, rolles theorem can be applied to $f$ on the interval $-2,1$.

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.\n\n$f(x)=(x - 1)^4-9(x - 1)^2+8;-2,1$\n\nsince $f$ is not continuous on the closed interval $-2,1$, not differentiable on the open interval $(-2,1)$, and $f(-2)=f(1)$, rolles theorem can be applied to $f$ on the interval $-2,1$.

Answer

Explanation:

Step1: Check continuity

A polynomial function (y = f(x)=(x - 1)^{4}-9(x - 1)^{2}+8) is continuous everywhere. So (f(x)) is continuous on the closed interval ([-2,1]).

Step2: Check differentiability

The derivative of (y=f(x)) using the chain - rule. If (u=x - 1), then (y = u^{4}-9u^{2}+8). The derivative (y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}). (\frac{dy}{du}=4u^{3}-18u) and (\frac{du}{dx}=1), so (f^\prime(x)=4(x - 1)^{3}-18(x - 1)). A polynomial function is differentiable everywhere. So (f(x)) is differentiable on the open interval ((-2,1)).

Step3: Check (f(-2)) and (f(1))

Calculate (f(-2)=(-2 - 1)^{4}-9(-2 - 1)^{2}+8=81-81 + 8=8). Calculate (f(1)=(1 - 1)^{4}-9(1 - 1)^{2}+8=8). So (f(-2)=f(1)).

Step4: Apply Rolle's Theorem

Since (f(x)) is continuous on ([-2,1]), differentiable on ((-2,1)) and (f(-2)=f(1)), by Rolle's Theorem, there exists (c\in(-2,1)) such that (f^\prime(c)=0). Set (f^\prime(x)=4(x - 1)^{3}-18(x - 1)=0). Factor out (2(x - 1)): (2(x - 1)[2(x - 1)^{2}-9]=0). Case 1: (x - 1=0) gives (x = 1) (but (1\notin(-2,1))). Case 2: (2(x - 1)^{2}-9=0), then ((x - 1)^{2}=\frac{9}{2}), (x-1=\pm\frac{3}{\sqrt{2}}), (x=1\pm\frac{3}{\sqrt{2}}). The value (x=1-\frac{3}{\sqrt{2}}\approx1 - 2.12=-1.12\in(-2,1)).

Answer:

Since (f) is continuous on the closed interval ([-2,1]), differentiable on the open interval ((-2,1)), and (f(-2)=f(1)), Rolle's Theorem can be applied. The point (c = 1-\frac{3\sqrt{2}}{2}) (or approximately (c=-1.12)) is guaranteed to exist by Rolle's Theorem.