$f(1)=-3$ and $ f'(x)\geq7$ how small is $f(5)$? Here is a test question in my textbook. 

Suppose :  $f(1)=-3$ and $f'(x)\geq7$ How small $f(5)$ can be possibly
  :
a)$25$
  b)$-21$
  c)$28$
  e)$31$
  f) None of others.

I just have this : because $f'(x)>0$ then $f(x)$ increasing. So, for all $x$ greater than $1$, $f(x)$ will greater than $-3$.
After this conclusion, I don't have any idea.
Because of the above conclusion, I post a,b,c. Because (I think) this problem will not have a fixed solution. So, we should  choose which solution is the most approximate.
 A: Hint 
$$\frac{f(5)-f(1)}{5-1} = f'(c) \geq 7 \,.$$
A: A previous answer uses an intuitive "let's use a linear function with a minimal derivative" solution.
For a rigorous argument, one can use the mean value theorem (in undergraduate mathematics, this is considered a "basic" theorem). This theorem says that, if $f$ is continuously differentiable*, then there must be an $x\in[1,5]$ (that is, an $x$ between 1 and 5) such that:
$$f'(x)=\frac{f(5)-f(1)}{5-1}$$
First note that I do not know what $x$ is. It is definitely in the range $[1,5]$, but the mean value theorem doesn't tell us what it is.
Also notice that I'm using a strict equality here - mathematicians like that, because we can now solve for $f(5)$ (which is good, to say the least):
$$f(5)=4f'(x)+f(1)$$
But now it is easy! If $f'(x)\geq7$ (for /any/ $x$), then $4f'(x)\geq28$, so $4f'(x)+f(1)\geq28+f(1)=25$, thus answer a) would be correct.
*There are many functions which are not continuously differentiable, but we usually do not consider them. Also, the question implicitly stated that the second derivative exists for all $x$ (namely, for all $x$ it has some value greater than or equal to 7), which is sufficient for the mean value theorem.
A: So you have a line of the form $y=7x+c$ , we take $7$ here because we want the minimum , if you take more then it will be more steep hence you get high value. 
To get $c$ you have initial condition , put $x=1$ and $y=-3$ to get $c=-10$ . 
Now you have $y=7x-10$ 
Put $x=5$ to get $y=25$
