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

Explanation:

Step1: Check 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 $[0,7]$.

Step2: Check differentiability

The derivative of $f(x)$ using the product - rule or by expanding first. If $f(x)=x^{3}-14x^{2}+49x$, then $f^\prime(x)=3x^{2}-28x + 49$. Polynomials are differentiable everywhere, so $f(x)$ is differentiable on $(0,7)$.

Step3: Check $f(0)$ and $f(7)$

Calculate $f(0)=0\times(0 - 7)^2=0$ and $f(7)=7\times(7 - 7)^2=0$.

Step4: Apply Rolle's theorem

Since $f(x)$ is continuous on $[0,7]$, differentiable on $(0,7)$ and $f(0)=f(7)=0$, Rolle's theorem applies. Set $f^\prime(x)=0$. We have $3x^{2}-28x + 49 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here $a = 3$, $b=-28$, $c = 49$. $x=\frac{28\pm\sqrt{(-28)^{2}-4\times3\times49}}{2\times3}=\frac{28\pm\sqrt{784 - 588}}{6}=\frac{28\pm\sqrt{196}}{6}=\frac{28\pm14}{6}$. $x_1=\frac{28 + 14}{6}=\frac{42}{6}=7$ and $x_2=\frac{28-14}{6}=\frac{14}{6}=\frac{7}{3}$. But $x = 7$ is an endpoint, so the value of $x$ in the open - interval $(0,7)$ is $x=\frac{7}{3}$.

Answer:

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