Question

If $f\left(0\right)=2.0$ and $f^{\prime }\left(x\right)=\frac{1}{5-x^2}$, then the lower and upper bounds of $f\left(1\right)$ estimated by the mean value theorem are

• A. $1.9,2.2$
• B. $2.2,225$
• C. $2.25,2.5$
• D. None of the above

Answer 1

The correct option is C $2.25,2.5$

It is given that $f\left(0\right)=2$ and we have to find f(1) so we can assume that $f\left(x\right)$ is defined in $[0,1]$
By lagrange's means value theorem
$\exists cϵ(0,1)$ such that
$f^{\prime }\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
$\Rightarrow f^{\prime }\left(c\right)=f\left(1\right)-2$
$\because$In $[0,1],f^{\prime }\left(x\right)=\frac{1}{5-x^2}>0$.
So $f\left(x\right)$ is an increasing function & $f"\left(x\right)=\frac{2x}{(5-x^2{)}^2}>0$.
So $f^{\prime }\left(x\right)$ is also an increasing funciton
$min{f}^{\prime }\left(c\right)\equiv \frac{1}{5-0}\equiv \frac{1}{5}$ and $max.f^{\prime }\left(c\right)=\frac{1}{5-1}=\frac{1}{4}$
Hence, $Min{f}^{\prime }\left(x\right)<f^{\prime }\left(c\right)<Max{f}^{\prime }\left(x\right)$
$\Rightarrow \frac{1}{5}<f\left(1\right)-2<\frac{1}{4}\Rightarrow 2.2<f\left(1\right)<2.25$