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 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 $-2,1$.\n\nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\n\na. rolles theorem applies, and the point(s) it guarantees to exist is/are $c=-1.12$\n(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\nb. rolles theorem cannot be applied.

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 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 $-2,1$.\n\nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\n\na. rolles theorem applies, and the point(s) it guarantees to exist is/are $c=-1.12$\n(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\nb. rolles theorem cannot be applied.

Answer

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\notin(-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})