Question

For what values of $a,m$ and $b$ Lagrange's mean value theorem is applicable to the function $f\left(x\right)$ for $x\epsilon [0,2]$
$f\left(x\right)=\{\begin{array}{cc}3& x=0\\ -x^2+a& 0<x<1\\ mx+b& 1\le x\le 2\end{array}$

• A. $a=3;m=2;b=0$
• B. $a=3;m=-2;b=4$
• C. $a=3;m=2;b=1$
• D. No such $a,m,b$ exist

Answer 1

The correct option is B $a=3;m=-2;b=4$

Using Lagrange mean value theorem

Let $f:[a,b]\to R$ be a continuous function and differentiable on $(a,b)$. Then there exists $c\in (a,b)$ such that $f\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Hence $f\left(x\right)$ is continuous

$f\left(0\right)=3=a$

$f\left(1\right)=2=m+b$

Hence $b=4$

$f\left(x\right)$ is differentiable at $1$

So $f^{\prime }\left(1\right)=-2=m$