Answer: c) [tex]\dfrac{3}{2}[/tex] .
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
[tex]\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}[/tex]
Given function : [tex]f(x) = x^2[/tex]
Interval : [0,3]
Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that
[tex]f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3[/tex]
[tex]\Rightarrow\ f'(c)=3\ \ \ ...(i)[/tex]
Since [tex]f'(x)=2x[/tex]
then, at x=c, [tex]f'(c)=2c\ \ \ ...(ii)[/tex]
From (i) and (ii), we have
[tex]2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}[/tex]
Hence, the correct option is c) [tex]\dfrac{3}{2}[/tex] .
