Question

Verify Lagrange's Mean Value Theorem for the function $f\left(x\right)=x^2+x-1$ in the interval $[0,4]$.

Answer 1

Here the function $f\left(x\right)=x^2+x-1$ is continuous in $[0,4]$ and differentiable on $(0,4)$, So,

then there will be existat least one real number $C$

here $0<c<4$ which satisfy

$F\left(c\right)=\frac{F\left(b\right)-F\left(a\right)}{b-a}$

So, $F\left(x\right)=x^2+x-1$

$F\left(x\right)=2x+1\Rightarrow F\left(c\right)=2c+1$

$2c+1=\frac{F\left(4\right)-F\left(0\right)}{4-0}=(4{)}^2+(4)-1$

$=\frac{\left\{\right(4{)}^2+\left(4\right)-1\}-\{0^2+0-1\}}{4}$

$2c+1=\frac{19+1}{4}$

$2c+1=5$

$2c=4$

$c=2,c\in (0,4)$