Let $f ( x ) = x + \frac { 1 } { x } f$ on $\left[ \frac { 1 } { 2 } , 2 \right] .$ Then the set of all c in $\left( \frac { 1 } { 2 } , 2 \right)$ guaranteed by the Mean Value Theorem is

A) $\left\{ \frac { 3 } { 4 } \right\}$
B) $\{ 1 \}$
C) $\left\{ \frac { 5 } { 4 } \right\}$
D) $\left\{ \frac { 3 } { 2 } \right\}$
E) $\left\{ \frac { 7 } { 4 } \right\}$

Question

Let $f ( x ) = x + \frac { 1 } { x } f$ on $\left[ \frac { 1 } { 2 } , 2 \right] .$ Then the set of all c in $\left( \frac { 1 } { 2 } , 2 \right)$ guaranteed by the Mean Value Theorem is

A) $\left\{ \frac { 3 } { 4 } \right\}$
B) $\{ 1 \}$
C) $\left\{ \frac { 5 } { 4 } \right\}$
D) $\left\{ \frac { 3 } { 2 } \right\}$
E) $\left\{ \frac { 7 } { 4 } \right\}$

Quizplus · Accepted Answer

The Mean Value Theorem states that if a function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$. For $f(x) = x + \frac{1}{x}$, we find $f'(x) = 1 - \frac{1}{x^2}$. To apply the Mean Value Theorem, we need to find a $c$ such that $f'(c) = \frac{f(2) - f(\frac{1}{2})}{2 - \frac{1}{2}}$. Calculating $f(2) = 2 + \frac{1}{2} = 2.5$ and $f(\frac{1}{2}) = \frac{1}{2} + 2 = 2.5$, we see that the numerator of the fraction is 0, which means $f'(c) = 0$. Solving $1 - \frac{1}{c^2} = 0$ gives $c = 1$, which is the only value in the interval $(\frac{1}{2}, 2)$ that satisfies the conditions of the Mean Value Theorem for this function.

Let f(x)=x+1xff ( x ) = x + \frac { 1 } { x } ff(x)=x+x1​f

Let $f ( x ) = x + \frac { 1 } { x } f$