Question

Examine if Mean value Theorem applies to $f\left(x\right)=x^3+3x^2-5x$ in the interval [1,2]. If it does, then find the intermediate point whose existence is asserted by theorem.

• A. Mean Value theorem is applicable and intermediate points are $c=-3.55,1.55$
• B. Mean Value theorem is not applicable
• C. Mean Value theorem is applicable and intermediate points are $c=3.55,-1.55$
• D. none of these

Answer 1

The correct option is A Mean Value theorem is applicable and intermediate points are $c=-3.55,1.55$

$f\left(x\right)=x^3+3x^2-5x$
Now $f\left(x\right)$ is continuous on [1,2] and differentiable on $(1,2)$.
$f\left(1\right)=-1$
$f\left(2\right)=10$
Hence by applying LMVT, we get
$\frac{f\left(2\right)-f\left(1\right)}{2-1}=f^{\prime }\left(c\right)$
$11=3c^2+6c-5$
$3c^2+6c-16=0$
$c=\frac{-6\pm \sqrt{36+3\left(64\right)}}{6}$
$=\frac{-6\pm 2\sqrt{9+48}}{6}$
$=\frac{-3\pm \sqrt{57}}{3}$
Hence
$c=1.5166$ and $c=-3.5166$
However 'c' has to be $1<c<2$
Hence the only possible value of c for LMVT to hold is 1.5166.