Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
f(x) = √x , [0, 25]
To find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval [0, 25] for the function f(x) = √x, we need to find a point in the interval where the instantaneous rate of change (the derivative) of the function is equal to the average rate of change (the slope of the secant line) on the interval.
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In this case, we have f(x) = √x on the interval [0, 25].
First, let's find f'(x), the derivative of f(x). Taking the derivative of √x using the power rule, we get:
f'(x) = (1/2) * x^(-1/2)
Next, let's find f(25) and f(0), which are the y-values of the endpoints of the interval. Plugging x = 25 and x = 0 into f(x), we get:
f(25) = √25 = 5
f(0) = √0 = 0
Now, let's find the slope of the secant line of f(x) on the interval [0, 25]:
Slope = (f(25) - f(0))/(25 - 0) = (5 - 0)/25 = 1/5
Finally, we need to find a value of c in the interval (0, 25) such that f'(c) = 1/5.
Setting f'(x) = 1/5, we have:
(1/2) * x^(-1/2) = 1/5
To solve this equation, we multiply both sides by 2:
x^(-1/2) = 2/5
Taking the reciprocal of both sides, we get:
x^(1/2) = 5/2
Squaring both sides, we have:
x = (5/2)^2 = 25/4
Therefore, the number c that satisfies the conclusion of the Mean Value Theorem on the interval [0, 25] for the function f(x) = √x is c = 25/4.
eh?
slope of line from 0,0 to 25,5= 5
...
I suspect the slope is 1/5
Now find a new c.
Not use full
No understand
f'= 1/(2sqrtx)
f(0)=0
f(25)=5
slope of line from 0,0 to 25,5= 5
so to satisfy the conclusion,
5= 1/(2sqrtx) if x exists on the domain.
sqrtx= 1/10
c= .01