Assuming [tex]f[/tex] is differentiable everywhere, then by the mean value theorem, there is some [tex]4<c<9[/tex] such that
[tex]f'(c)=\dfrac{f(9)-f(4)}{9-4}\implies 5f'(c)=f(9)-f(4)[/tex]
Since [tex]4\le f'(x)\le5[/tex],
[tex]\implies f(9)-f(4)=5f'(c)\le5\cdot5=25[/tex]
[tex]\implies f(9)-f(4)=5f'(c)\ge5\cdot4=20[/tex]
So [tex]20\le f(9)-f(4)\le25[/tex].
