Consider the function f(x)=6sqrt(x)+1 on the interval [3,5] . Find the average or mean slope of the function on this interval. I FOUND THIS: (6sqrt(5)-6sqrt(3))/2
By the Mean Value Theorem, we know there exists a c in the open interval (35) such that f(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
I NEED C
PLEASE HELP
differentiate f(x) -> f'(x)=3/sqrt(x)(double check) and supposingly that has at least one value equal to mean slope:(6sqrt(5)-6sqrt(3))/2 hence you want to solve: 3/sqrt(x) = (6sqrt(5)-6sqrt(3))/2.
Solve this and you will get x=some value. Put the x=some value back to the original function to get y and you get C (x,y).
Can you please give me the answer to c. I cannot figure it out.
this is homework help, not homework solve. Ask for help is ok, but no one guarrantees to solve your problems. It's your problem after all.
To find the value of c that satisfies the mean value theorem, you need to consider the derivative of the function f(x) = 6√(x) + 1.
First, find the derivative of f(x) with respect to x. The derivative of 6√(x) + 1 can be found by using the power rule and the chain rule:
f'(x) = (d/dx) [6√(x)] + (d/dx) [1]
= 6(1/2)(1/x^0.5) + 0
= 3/x^0.5
Next, calculate the average slope or mean slope of the function f(x) over the interval [3, 5]. The mean slope is given by the formula:
Mean slope (m) = [f(b) - f(a)] / (b - a)
Where a = 3 and b = 5.
Plugging in the values into the equation, we get:
m = [f(5) - f(3)] / (5 - 3)
= [6√(5) + 1 - (6√(3) + 1)] / 2
= 6√(5) - 6√(3) / 2
So, as you mentioned, the mean slope of the function on the interval [3, 5] is (6√(5) - 6√(3)) / 2.
Now, we want to find the value of c that satisfies the mean value theorem. The mean value theorem states that there exists a c in the open interval (3, 5) such that f'(c) = m.
Since f'(x) = 3/x^0.5, set it equal to the mean slope:
3/c^0.5 = (6√(5) - 6√(3)) / 2
To solve for c, cross-multiply and simplify:
2 * 3 = (6√(5) - 6√(3)) * c^0.5
6 = 6√(5)c^0.5 - 6√(3)c^0.5
Rearrange the equation:
6√(3)c^0.5 = 6√(5)c^0.5 - 6
Add 6 to both sides:
6√(3)c^0.5 + 6 = 6√(5)c^0.5
Divide both sides by 6 and simplify:
√(3)c^0.5 + 1 = √(5)c^0.5
Subtract 1 from both sides:
√(3)c^0.5 = √(5)c^0.5 - 1
Square both sides to eliminate the square root:
3c = 5c - 2√(5)c^0.5 + c
Rearrange the equation:
-2√(5)c^0.5 = -2c + 2√(5)c^0.5
Combine like terms:
4√(5)c^0.5 = 2c
Divide both sides by 2c to isolate the square root:
2√(5)c^(-0.5) = 1
Square both sides to get rid of the square root again:
20c^(-1) = 1
Rearrange the equation:
c^(-1) = 1/20
Take the reciprocal to solve for c:
c = 20
Therefore, the value of c that satisfies the mean value theorem for the given function on the interval [3, 5] is c = 20.