Answer:
This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3
Step-by-step explanation:
The given function is
When we differentiate this function with respect to x, we get;
We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;
![c-3=\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c-3%3D%5Csqrt%5B3%5D%7B63.15789%7D)
![c=3+\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c%3D3%2B%5Csqrt%5B3%5D%7B63.15789%7D)
If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).
But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.
Answer:
Step-by-step explanation:
From the table attached,
x-intercept of the linear function is, the value of 'x' when f(x) = 0
x = 3 [x-intercept]
Function 'g' is the sum of 2 and the cube root of the sum of three time x and 1.
g(x) = ![2+\sqrt[3]{3x+1}](https://tex.z-dn.net/?f=2%2B%5Csqrt%5B3%5D%7B3x%2B1%7D)
For x-intercept,
g(x) = 0
![2+\sqrt[3]{3x+1}=0](https://tex.z-dn.net/?f=2%2B%5Csqrt%5B3%5D%7B3x%2B1%7D%3D0)
3x + 1 = (-2)³
3x + 1 = -8
3x = -8 - 1
3x = -9
x = -3
Therefore, the x-intercept of function 'f' is different or greater than the x-intercept of function g.
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Please check the attached answer of picture for explanation.
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Answer:
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Step-by-step explanation:
