Let $f:[0, 1] \to [a, b]$ be a continuous function. Then there exists $x ∈ [0, 1]$ such that $f(x) = (b − a)x + a$. 
Let $f:[0,1] \to [a,b]$ be a continuous function.  Then there exists $x \in [0,1]$ such that $f(x) = (b-a)x + a$.

I see that the line that the statement proposes to exist would pass through the points $(0)=$ and $(1)=$ so I want to use the intermediate value theorem which says that a function continuous on an interval takes on all values between any two of its values.
Graphically I can intuit that this's happening in this situation but I don't know how to express it formally.
 A: NB:  The goal of this answer is to be pedagogical.  The exposition describes the way in which I might think about this problem.  It is, therefore, wordy.  A more concise summary is given at the end.

The intermediate value theorem (IVT) states

Theorem:  Suppose that $f$ is continuous on the interval $[a,b]$, and that $$y \in (\min\{f(a), f(b)\}, \max\{f(a),f(b)\}).$$  Then there exists some $x \in [a,b]$ such that $f(x) = y$.

In other words, a continuous function on a closed interval attains every value between the values at its endpoints.  In order to apply the IVT to the given problem, we need three ingredients:

*

*a continuous function on some interval (the interval in question is probably $[0,1]$ in this case),

*the values of that function at the endpoints; and

*some value which you can know, for sure, is between those end values.

Often, it is useful to start with (2).  In this particular case, a reasonable starting place might be to rearrange the desired equality:
$$ f(x) = (b-a)x + a \iff f(x) - (b-a)x - a = 0. $$
The expression on the left-hand side of this equation is a function, and we are hoping to find some input for this function which outputs zero.  This seems to resolve (2), assuming that the remaining hypotheses of the IVT are satisfied.
For (1), it is necessary to know that this function on the left-hand side is continuous on $[0,1]$.  For brevity, define
$$ g(x) = f(x) - (b-a)x - a. $$
Note that $g$ is the sum of two functions:  $f$, which is continuous on $[0,1]$ by hypothesis; and the function $x \mapsto (b-a)x + a$, which is continuous on all of $\mathbb{R}$ (and, in particular, continuous on $[0,1]$.  The sum of two continuous functions is continuous, hence $g$ is continuous.  This checks off (1).
Finally, we need to know what this function does on the endpoints.  Observe that
$$ g(0) = f(0) - (b-a)(0) - a = f(0) - a
\qquad\text{and}\qquad
g(1) = f(1) - (b-a)(1) - a = f(1) - b. $$
Now, if we could show that one of these two quantities is positive, and the other negative, we would be done.  Unfortunately, it is not "obvious" that this is the case.  Indeed, I'm a little stuck here.
So... let's see if there is any hypothesis which we haven't used yet. Let's reread the statement of the problem:

Let $f : [0,1] \to [a,b]$ be a continuous function.  Then there exists some $x \in [0,1]$ such that $f(x) = (b-a)x + a$.

We have not yet used the fact that the codomain of $f$ is $[a,b]$ (that is, the image (or range) of $f$ lies in that interval).  In particular, this means that
$$ a \le f(x) \le b$$
for all $x \in [0,1]$.  But then
$$ g(0) = f(0) - a \ge a-a = 0, $$
since $f(0) \ge a$.  Similarly,
$$ g(1) = f(1) - b \le b-b = 0. $$
Aha!  It looks like we can guarantee that $g(0)$ is positive, and that $g(1)$ is negative.
Almost...
There is a pesky little issue, in that the inequalities are weak—it is possible that either of those inequalities might be equalities, which would mess up the desired application of the intermediate value theorem.  So let's make sure that this isn't really a problem, and handle these two cases on their own.
Observe that if $g(0) = 0$, then we get
$$ 0 = g(0) = f(0) - (b-a)(0) - a
\iff f(0) = (b-a)(0) + a, $$
which is the desired result (with $x=0$).  Something similar happens if $g(1) = 0$.
Great!  There is no problem if either $g(0) = 0$ or $g(1) = 1$.
So assume that
$$ g(1) < 0 < g(0). $$
We have already concluded that $g$ is continuous on $[0,1]$, and we have just discovered that we can assume that $g(1)$ is negative and that $g(0)$ is positive.  Therefore, by the IVT, there must be some $x \in [a,b]$ such that $g(x) = 0$.  But for this $x$,
$$ 0 = g(x) = f(x) - (b-a)x - a
\iff f(x) = (b-a)x + a, $$
which is the desired result.

Taking my thinking, above, and writing it up cleanly, I might produce something like the following:

Problem: Show that if $f : [0,1] \to [a,b]$ is continuous, then there exists some $c \in [0,1]$ such that
$$ f(c) = (b-a)c + a. $$
Solution:  Let $g$ be defined on $[0,1]$ by the formula
$ g(x) = f(x) - (b-a)x - a. $
Observe that
$$ g(1) = f(1) - (b-a)(1) - a = f(1) - b \le 0, $$
since $f(x) \le b$ for all $x \in [0,1]$.  By a similar argument,
$$ g(0) = f(0) - (b-a)(1) - a = f(0) - a \ge 0. $$
If $g(1) = 0$, then $ f(1) = (b-a)(1) + a$, so take $c=1$.  If $g(0) = 0$, then $f(0) = (b-a)(0) + a$, so take $c=0$.  Otherwise,
$$ g(1) < 0 < g(0). $$
But $g$ is the sum of two continuous functions, and is therefore continuous itself.  Hence, by the intermediate value theorem, there exists some $c \in [0,1]$ such that $g(c) = 0$.  But then
$$ f(c) = (b-a)c + a, $$
which completes the argument.

