Question

the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2 e) None of the above

Answer 1

Answer:  c) $\dfrac{3}{2}$ .

Step-by-step explanation:

Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

$\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}$

Given function : $f(x) = x^2$

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

$f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3$

$\Rightarrow\ f'(c)=3\ \ \ ...(i)$

Since $f'(x)=2x$

then, at x=c, $f'(c)=2c\ \ \ ...(ii)$

From (i) and (ii), we have

$2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}$

Hence, the correct option is c) $\dfrac{3}{2}$ .