**General Solution Method**

The Mean Value Theorem has two hypotheses:

**H1** : #f# is continuous on the closed interval #[a,b]#

**H2** : #f# is differentiable on the open interval #(a,b)#.

In this question, #f(x) = 4x# , #a=4# and #b=9#.

This function is a polynomial, so it is continuous on its domain, #(-oo, oo)#, so it is continuous on #[11, 23]#

#f'(x)= 4# which exists for all #x# (for all #x in (4,9)#,

So #f# is differentiable on #(4, 9)#

Therefore the hypotheses of The Mean Value Theorem are true for this function on this interval.

In general, to find #c#, solve #f'(x) = (f(b)-f(a))/(b-a)#, keeping only the solutions in #(a,b)#.

In this question we need to solve #4 = (4(9)-4(4))/(9-4) = 20/5 = 4#.

Every real number is a solution, so for #c# we may take any number in #(4,9)#.

**Alternative solution for a linear function** (like this one).

#f(x) = 4x# is a linear function, so the secant line joining the endpoints on the interval is exactly the graph of the function. It has slope #4#.

The #c# is the value at which the tangent line has the same slope as the secant line joining the endpoints. But at every number, the tangent line has slope #4#, so any number in the open interval will work for #c#.