Answer:
Step-by-step explanation:
DE given is
y''+8y'+15y=0, y(0)=0, y'(0)=1
Take Laplace on the DE
We get

Simplify to get
Y(s) = ![\frac{1}{2}[ {\frac{1}{s+3} -\frac{1}{s+5} }]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%20%7B%5Cfrac%7B1%7D%7Bs%2B3%7D%20-%5Cfrac%7B1%7D%7Bs%2B5%7D%20%7D%5D)
Take inverse

Answer:
Domain: 1 ≤ x ≤ 4
Range : 1 ≤ f(x) ≤ 4
Step-by-step explanation:
The domain of a function f(x) is the limit within which the values of x varies.
Here, in the graph, it shows that the maximum value of x is 4 and the minimum value of x is 1.
Therefore, the domain of the function is 1 ≤ x ≤ 4
Again the range of a function f(x) is the limit within which the values of f(x) vary.
Here, the graph shows that the maximum value of f(x) is 4 and the minimum value of f(x) is 1.
Therefore, the range of the function f(x) is 1 ≤ f(x) ≤ 4. (Answer)
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:
www.mahtpapa.com
Step-by-step explanation:
Answer:
9x^9-18
Step-by-step explanation: