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)= sqrt of x [0,9]
c =

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.)

table a is wierd

step-by-step explanation:

6*f