Respuesta :
Answer:
9/4
Step-by-step explanation:
f(x) is continuous and differentiable on (0,9).
We want to find c using the following equation.
f'(c)=(f(9)-f(0))/(9-0)
This will require us to find f'(x) first.
f(x)=sqrt(x) is the same as f(x)=(x)^(1/2)
Using power rule to differentiate this gives f'(x)=(1/2)(x)^(1/2-1) or simplified f'(x)=(1/2)x^(-1/2) or f'(x)=1/(2x^(1/2)).
So we want to solve:
(1/2)c^(-1/2)=(f(9)-f(0))/(9-0)
Simplify denominator on right:
(1/2)c^(-1/2)=(f(9)-f(0))/9
This will require us to find f(9) and f(0).
If f(x)=sqrt(x), then f(9)=sqrt(9)=3 and f(0)=sqrt(0)=0.
So we have the following equation so far:
(1/2)c^(-1/2)=(3-0)/9
Simplify numerator on right:
(1/2)c^(-1/2)=3/9
Multiply both sides by 2:
c^(-1/2)=6/9
Raise both sides to the -2 power:
c^(1)=(6/9)^(-2)
Note c^1=c:
c=(6/9)^(-2):
Note negative exponent means to find reciprocal of base to change exponent to opposite
c=(9/6)^2
Apply the second power:
c=81/36
Reduce by dividing top and bottom by 9:
c=9/4
This means the slope of the tangent to the curve f at x=9/4 is the same value as the slope of the secant line going through points (0,0) and (9,3).
Also 9/4 is between 0 and 9... According to the theorem we were suppose to get a value c between x=0 and x=9.
Confirmation:
Slope of the secant line is (3-0)/(9-0)=3/9=1/3.
Slope of the tangent line to curve f at x=9/4.
f'(x)=(1/2)x^(-1/2)
f'(9/4)=(1/2)(9/4)^(-1/2)
f'(9/4)=(1/2)(3/2)^(-1)
f'(9/4)=(1/2)(2/3)
f'(9/4)=1/3
They are indeed equal values (talking about the 1/3 from the secant and the tangent.)
