Question

The value of c in Mean value theorem for the function $f\left(x\right)=x^2$ in [2, 4] is

• A. $4$
• B. $2$
• C. $\frac{7}{2}$
• D. $3$

Answer 1

Answer: (D)

$3$

Mean value theorem states that if f(x) is defined and continuous on interval [a,b] and diffrentiable on (a,b)then there is atleast one number in c in interval (a,b) i.e. a<c<b such that
$f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
Graph of $f\left(x\right)=x^2$
$\therefore$Function is continuous
$f^{{}^{\prime }}\left(x\right)=2x$
$f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
$f^{{}^{\prime }}\left(c\right)=\frac{{\left(4\right)}^2-{\left(2\right)}^2}{4-2}$
$f^{{}^{\prime }}\left(c\right)=\frac{16-4}{2}=6$
$2x=6$
$x=3$
$\therefore c=3$ and $2<3<4$