Answer:
See Below.
Step-by-step explanation:
We want to show that the function:

Increases for all values of <em>x</em>.
A function is increasing whenever its derivative is positive.
So, find the derivative of our function:
![\displaystyle f'(x) = \frac{d}{dx}\left[e^x - e^{-x}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5Be%5Ex%20-%20e%5E%7B-x%7D%5Cright%5D)
Differentiate:

Simplify:

Since eˣ is always greater than zero and e⁻ˣ is also always greater than zero, f'(x) is always positive. Hence, the original function increases for all values of <em>x.</em>
Answer:
uhhh ok
Step-by-step explanation:
Answer:
t=3.2c
Step-by-step explanation:
After a little manipulation, the given diff'l equation will look like this:
e^y * dy = (2x + 1) * dx.
x^2
Integrating both sides, we get e^y = 2------- + x + c, or e^y = x^2 + x + c
2
Now let x=0 and y = 1, o find c:
e^1 = 0^2 + 0 + c. So, c = e, and the solution is e^y = x^2 + x + e.