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

The value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is 3/2.

Step-by-step explanation:

The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in (a, b) such that the average rate of change of the function over the interval [a, b] is equal to the instantaneous rate of change at c. In this case, the function is V(x) = x² and the interval is [0, 3].

To find the value of c, we need to determine the average rate of change and the instantaneous rate of change. The average rate of change is given by (V(3) - V(0))/(3 - 0) = (9 - 0)/3 = 3. The instantaneous rate of change can be found by taking the derivative of V(x) and evaluating it at c. Since V(x) = x², the derivative is V'(x) = 2x. Setting V'(c) = 2c equal to the average rate of change, we have 2c = 3. Solving for c, we find c = 3/2.

Answer 2

Answer: c)
`(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) = (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)=(f(3)-f(0))/(3-0)\\\\=(3^2-0^2)/(3)=(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=(3)/(2)`

Hence, the correct option is c)
`(3)/(2)` .