Proving an inequality involving norms of real functions. If $ f : [a,b] \subset \mathbb{R} \rightarrow \mathbb{R} $ is continuous and differentiable in $(a,b)$, then one can define a norm for such functions as
$$ \|f\| = |f(a)| + \max_{x \in (a,b)} |f ^\prime(x)| $$
I have already proven that it is in fact a norm.
Now I am trying to prove that this inequality holds.
$$ \max_{x \in [a,b]} |f(x)| \leq (1+b-a)\|f\| $$
There is a hint that I should remember that $f(x) = f(a) +\int_{a}^{x}f^\prime(t) dt$
I have tried several approaches with no success.
If I let $\theta = \arg\max|f(x)|$ and $\lambda = \arg\max|f^\prime(x)|$
Then I could use the hint and say that
$$ \|f\| = \|f(a) + \int_{a}^{x}f^\prime(t) dt \| \leq \|f(a)\| + \|\int_{a}^{x}f^\prime(t) dt \| $$
and I could get that that
$$ 0 \leq \|f\| \leq |f(\theta)| + |f^\prime(\lambda)|  $$
but now I have  $|f(\theta)| $ on the wrong side of the inequality and I can't find a way to move things around so that I could get any close to my goal.
I tried to define a function $g(x) = (1+b-a)f(x)$, and then I would only have to show that $|f(\theta)| \leq \|g\|$, but I can't find a way to do so.
Any hint would be greatly appreciated; specially about how to use the hint I already have.
 A: We have$$\left|f\left(x\right)\right|=\left|f\left(a\right)+\int_{a}^{x}f'\left(t\right)dt\right|\leq\left|f\left(a\right)\right|+\int_{a}^{x}\left|f'\left(t\right)\right|dt\leq\left|f\left(a\right)\right|+\max_{t\in\left(a,b\right)}\left|f'\left(t\right)\right|\left(b-a\right)\leq\left(1+b-a\right)\left|f\left(a\right)\right|+\max_{t\in\left(a,b\right)}\left|f'\left(t\right)\right|\left(1+b-a\right)=\left(1+b-a\right)\left\Vert f\right\Vert$$ 
 because $b-a>0$
  and $1<1+b-a$.
  So$$\max_{x\in\left[a,b\right]}\left|f\left(x\right)\right|\leq\left(1+b-a\right)\left\Vert f\right\Vert.$$
A: Here is another approach:
Using the MVT, we have that for any $x$ we can find a $c$ such that 
$$f(x)-f(a)=f'(c)(x-a).$$
Rearranging gives:
$$f(x)=f(a)+f'(c)(x-a).$$
Taking absolute values and using the triangle inequality
$$|f(x)|=|f(a)+f'(c)(x-a)|\leq |f(a)|+|f'(c)|\cdot|(x-a)|$$
then taking the max of everything:
$$\begin{aligned}
\max|f(x)|&\leq |f(a)|+\max|f'(c)|\cdot\max|(x-a)|\\
&\leq|f(a)|+\max|f'(x)|\cdot (b-a) \\
\end{aligned}$$
This is a much stronger condition. So we add to the right side $|f(a)|(b-a)+\max|f'(x)|$ to get 
$$
\begin{aligned}
\max|f(x)|&\leq|f(a)|+\max|f'(x)| (b-a)+|f(a)|(b-a)+\max|f'(x)|\\
 &=(|f(a)|+\max|f'(x)|) (1+b-a) \\
&=||f||(1+b-a) .
\end{aligned}
$$
