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(x - 7)^2; 0,7
select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. rolles theorem applies and the point(s) guaranteed to exist is/are c=
(type exact answer(s). use a comma to separate answers as needed)
b. rolles theorem does not apply.

Explanation:

Step1: Check the conditions of Rolle's Theorem

• Continuity: The function \(f(x)=x(x - 7)^{2}=x(x^{2}-14x + 49)=x^{3}-14x^{2}+49x\) is a polynomial. Polynomials are continuous everywhere, so \(f(x)\) is continuous on \([0,7]\).
• Differentiability: The derivative of \(f(x)\) using the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (where \(u = x\) and \(v=(x - 7)^{2}\)) or by differentiating \(f(x)=x^{3}-14x^{2}+49x\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=3x^{2}-28x + 49\). Polynomials are differentiable everywhere, so \(f(x)\) is differentiable on \((0,7)\).
• \(f(0)\) and \(f(7)\):
• \(f(0)=0\times(0 - 7)^{2}=0\).
• \(f(7)=7\times(7 - 7)^{2}=0\).

Since \(f(x)\) is continuous on \([0,7]\), differentiable on \((0,7)\) and \(f(0)=f(7)\), Rolle's Theorem applies.

Step2: Find \(c\) such that \(f^\prime(c)=0\)

Set \(f^\prime(x)=3x^{2}-28x + 49 = 0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 3\), \(b=-28\), \(c = 49\)).

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We get two solutions:

• \(x=\frac{28 + 14}{6}=\frac{42}{6}=7\) (but \(7\) is an endpoint of the interval \((0,7)\)).
• \(x=\frac{28-14}{6}=\frac{14}{6}=\frac{7}{3}\)

Answer:

A. Rolle's Theorem applies and the point(s) guaranteed to exist is/are \(c = \frac{7}{3}\)