Question

Verify Lagrange's mean value thorem for the function $f\left(x\right)=x{\left(x+4\right)}^2$ in $[0,4]$.

Answer 1

Legrange's Mean value theorem states that if a function $f\left(x\right)$ is continuous in $[a,b]$ and differentiable in $(a,b)$ then there is $c\in (a,b)$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here $f\left(x\right)=x(x+4{)}^2=x(x^2+16+8x)=x^3+8x^2+16x$

$f^{\prime }\left(x\right)=3x^2+16x+16$

$f\left(x\right)$ is continuous in $[0,4]$ and differentiable in $(0,4)$

$f\left(a\right)=f\left(0\right)=0$ and $f\left(4\right)=4^3+8(4{)}^2+16(4)=256$

$\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{256}{4}=64$

$f^{\prime }\left(c\right)=3c^2+16c+16=64$

$⟹3c^2+16c-48=0$

$c=\frac{-16\pm 8\sqrt{1}3}{6}=\frac{-8\pm 4\sqrt{1}3}{3}\approx 2.14,-7.474$

$2.14$ lies in $(0,4)$