Question

a. determine whether the mean value theorem applies to the function ( f(x)=1 + x^{2} ) on the interval ( -1,2 ).
b. if so, find the point(s) that are guaranteed to exist by the mean value theorem.
a. choose the correct answer below.
a. yes, because the function is continuous on the interval ( -1,2 ) and differentiable on the interval ( (-1,2) ).
b. no, because the function is differentiable on the interval ( (-1,2) ), but is not continuous on the interval ( -1,2 ).
c. no, because the function is continuous on the interval ( -1,2 ), but is not differentiable on the interval ( (-1,2) ).
d. no, because the function is not continuous on the interval ( -1,2 ), and is not differentiable on the interval ( (-1,2) ).
b. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the point(s) is/are ( x = ).
(simplify your answer. use a comma to separate answers as needed.)
b. the mean value theorem does not apply in this case.

Explanation:

Step1: Check continuity and differentiability

A polynomial function \(y = f(x)=1 + x^{2}\) is continuous everywhere (since \(y = ax^{n}+bx^{m}+\cdots\) with \(a,b,\cdots\in R\) and \(n,m,\cdots\) non - negative integers is continuous for all \(x\in R\)). So \(f(x)\) is continuous on \([-1,2]\).
The derivative of \(f(x)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\) is \(f^\prime(x)=2x\), which exists for all \(x\in R\). So \(f(x)\) is differentiable on \((-1,2)\).

Step2: Apply the Mean Value Theorem formula

The Mean Value Theorem states that \(f^\prime(c)=\frac{f(b)-f(a)}{b - a}\), where \(a=-1\), \(b = 2\).
First, find \(f(a)\) and \(f(b)\):
\(f(-1)=1+(-1)^{2}=2\), \(f(2)=1 + 2^{2}=5\).
Then \(\frac{f(2)-f(-1)}{2-(-1)}=\frac{5 - 2}{3}=1\).
Since \(f^\prime(x)=2x\), set \(f^\prime(c)=2c\) equal to \(\frac{f(2)-f(-1)}{2-(-1)}\).
We have the equation \(2c = 1\).

Step3: Solve for \(c\)

Solving \(2c=1\) gives \(c=\frac{1}{2}\).

Answer:

a. A. Yes, because the function is continuous on the interval \([-1,2]\) and differentiable on the interval \((-1,2)\)
b. A. The point(s) is/are \(x=\frac{1}{2}\)