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erastovalidia [21]
3 years ago
6

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
Mathematics
1 answer:
lilavasa [31]3 years ago
4 0

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} .

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