Find an Example of Function Where... 
*

*A function $f(x)$ is cts on $[0,10]$, takes the max value 5 and min value of -5 in this interval and $\int_0^{10} f(x) dx = 25$
-----What I'm Thinking: since its a cts function, i should try to avoid piece wise functions as they would probably not be cts. $f(x)$ has got to cross the x axis at some point. other than that i'm stumped.

*A differential equation that has solution $y = 3^x$
----- I'm mostly confused what it means to have a solution $y=3^x$.
 A: *

*Of course you can use a piecewise defined function here as long as you check that it is continuous everywhere and I think it is also the easiest way to do it like that.
Here you can define $f$ the following way: $f(x)=5$ for $x \geq 5$. This ensures that $\int_5^{10}f(x)dx = 25$. Now you can define the function on the other part of the interval in such a way that the integral vanishes. The most easiest way is to accomplish it by defining $f$ on this interval in an odd-style manner:
$$f(x) = 
\begin{cases}
5, \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \leq x  \leq 10 \\
-5+2x, \ 0 \leq x < 5\\
\end{cases}$$


*This means that you are looking for a differential equation $y'=f(y,x)$ (could also be of higher order, but I assume a first-order ODE is meant). Well, here you could just derivate the function $y$. So $$y(x) = 3^x = e^{\ln(3)x}$$ and therefore by the chain rule $$y'(x) = e^{\ln(3)x}\ln(3) = y(x)\ln(3).$$ So a possible differential equation would be $$y'(x) = y(x)\ln(3)$$
A: It's possible to do (a) without a piecewise function but it takes a bit more algebra. Let $f(x)$ be a quadratic passing $(0,-5)$: $f(x)=ax^2+bx-5$. The maximum of $5$ is at $x=-b/2a$, so we can solve for $b$. When we set the integral over $[0,10]$ to $25$ and solve for $a$, we find $f(x)=-\frac{9}{40}x^2+3x-5$. 
