Answer:
f(5) cannot exceed 30.
Step-by-step explanation:
f is differentiable in all R, so we can apply the Mean Value Theorem on the interval [0,5], since f is continuous on [0,5] and differentiable on (0,5).
By the MVT, there exists some number c∈(0,5) such that [tex]f(5)-f(0)=f'(c)(5-0)[/tex]. Substituting the value of f(0), this last equation is equivalent to [tex]f(5)+5=5f'(c)[/tex]. But f'(x)≤7 for all values of x, in particular for x=c, so applying this bound we obtain that [tex]f(5)+5=5f'(c)\leq5\cdot 7=35 [/tex]. Then [tex]f(5)+5=\leq35[/tex], so substracting 5 from both sides we conclude that [tex]f(5)=\leq30[/tex].
