Consider the interval (0, 5). Assuming [tex]f[/tex] is differentiable for all [tex]x\in(-5,5)[/tex], we know by the mean value theorem that there is some [tex]c\in(0,5)[/tex] such that
[tex]f(c)=\dfrac{f(5)-f(0)}{5-0}[/tex]
and given that [tex]5\le f'(x)\le6[/tex] and [tex]f(0)=7[/tex], we have
[tex]5\le\dfrac{f(5)-7}5\le6\implies32\le f(5)\le37[/tex]
