let $f(x) = \frac{1}{|x|}$.

can we use the mean value theorem to say the equation $f(x) = -\frac{1}{4}$ has a solution where $2

Question

let $f(x) = \frac{1}{|x|}$.

can we use the mean value theorem to say the equation $f(x) = -\frac{1}{4}$ has a solution where $2 < x < 4$?

choose 1 answer:

a  no, since the function is not differentiable on that interval.

b  no, since the average rate of change of $f$ over the interval $2 \leq x \leq 4$ isnt equal to $-\frac{1}{4}$.

c  yes, both conditions for using the mean value theorem have been met.

Accepted Answer

# Brief Explanations:
To apply the Mean Value Theorem (MVT), a function must be continuous on $[a,b]$ and differentiable on $(a,b)$. For $f(x)=\frac{1}{|x|}$, on the interval $(2,4)$, $x>0$, so $f(x)=\frac{1}{x}$ (since $|x| = x$ for $x>0$). The function $y = \frac{1}{x}$ is continuous on $[2,4]$ and differentiable on $(2,4)$ (derivative $f'(x)=-\frac{1}{x^2}$). Now, check the average rate of change over $[2,4]$: $\frac{f(4)-f(2)}{4 - 2}=\frac{\frac{1}{4}-\frac{1}{2}}{2}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}$? Wait, no—wait, the question is about $f'(x)=-\frac{1}{4}$. Wait, no, let's recalculate: $f(4)=\frac{1}{4}$, $f(2)=\frac{1}{2}$. So $\frac{f(4)-f(2)}{4 - 2}=\frac{\frac{1}{4}-\frac{1}{2}}{2}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}$? But wait, the MVT says there exists $c$ in $(2,4)$ such that $f'(c)=\frac{f(4)-f(2)}{4 - 2}$. Wait, but the question is about $f'(x)=-\frac{1}{4}$. Wait, no, maybe I made a mistake. Wait, $f(x)=\frac{1}{|x|}$, for $x>0$, $f(x)=\frac{1}{x}$, derivative $f'(x)=-\frac{1}{x^2}$. Let's check if the average rate of change is $-\frac{1}{4}$. Wait, $\frac{f(4)-f(2)}{4 - 2}=\frac{\frac{1}{4}-\frac{1}{2}}{2}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}$, which is not $-\frac{1}{4}$. Wait, but the options: Option B says "No, since the average rate of change... isn't equal to $-\frac{1}{4}$". Wait, but wait, the MVT requires that the average rate of change (slope of secant) equals $f'(c)$ for some $c$. So if we want to say there's a $c$ with $f'(c)=-\frac{1}{4}$, we need the average rate of change to be $-\frac{1}{4}$. Let's calculate the average rate of change: $\frac{f(4)-f(2)}{4 - 2}=\frac{\frac{1}{4}-\frac{1}{2}}{2}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}$, which is not $-\frac{1}{4}$. So the average rate of change over $[2,4]$ is $-\frac{1}{8}$, not $-\frac{1}{4}$. Therefore, the reason is that the average rate of change isn't equal to $-\frac{1}{4}$, so we can't use MVT to say $f'(x)=-\frac{1}{4}$ has a solution? Wait, no—wait, the MVT says that $f'(c)$ equals the average rate of change. So if the average rate of change is not $-\frac{1}{4}$, then there's no $c$ with $f'(c)=-\frac{1}{4}$ by MVT. Wait, but let's check the function's differentiability: for $x>0$, $f(x)=\frac{1}{x}$ is differentiable on $(2,4)$ (since derivative exists there). So the function is differentiable on $(2,4)$ (because for $x>0$, $|x|=x$, so $f(x)=\frac{1}{x}$, which is differentiable). Wait, so why is option A wrong? Wait, no—wait, $f(x)=\frac{1}{|x|}$ is differentiable on $(2,4)$ (since $x>0$ there, so $|x|=x$, and $\frac{1}{x}$ is differentiable). So the problem is that the average rate of change over $[2,4]$ is $-\frac{1}{8}$, not $-\frac{1}{4}$, so the MVT can't guarantee a solution for $f'(x)=-\frac{1}{4}$ because the average rate of change isn't equal to $-\frac{1}{4}$. So option B is correct? Wait, but wait, let's re-express: The MVT states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b - a}$. So to use MVT to say $f'(x)=k$ has a solution, $k$ must equal $\frac{f(b)-f(a)}{b - a}$. Here, $k = -\frac{1}{4}$, and $\frac{f(4)-f(2)}{4 - 2}=\frac{\frac{1}{4}-\frac{1}{2}}{2}=-\frac{1}{8}
eq-\frac{1}{4}$. Therefore, the average rate of change isn't equal to $-\frac{1}{4}$, so we can't use MVT to say $f'(x)=-\frac{1}{4}$ has a solution. So option B is correct. Wait, but earlier I thought maybe option A, but no—for $x>0$, $f(x)=\frac{1}{x}$ is differentiable on $(2,4)$ (since $x$ is never zero there, and $\frac{1}{x}$ is differentiable for $x
eq0$). So the function is differentiable on $(2,4)$ (because $x\in(2,4)$ implies $x>0$, so $|x|=x$, $f(x)=\frac{1}{x}$, which is differentiable). So option A is wrong (since it is differentiable on $(2,4)$). Option C is wrong because the average rate of change isn't $-\frac{1}{4}$, so the MVT doesn't guarantee a solution for $f'(x)=-\frac{1}{4}$ (it guarantees a solution for $f'(x)=$ average rate of change, which is $-\frac{1}{8}$ here). Therefore, option B is correct: "No, since the average rate of change of $f$ over the interval $2\leq x\leq4$ isn't equal to $-\frac{1}{4}$".
# Answer:
B. No, since the average rate of change of $ f $ over the interval $ 2 \leq x \leq 4 $ isn't equal to $ -\frac{1}{4} $.

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