Sketch means to draw a picture describing it. Not to actually find the expression of one.
Assuming continuity and differentiability then:
An absolute maximum at $2$ means: as $1 < 2 < 5$ it must also be a local maximum so the function "plateaus" at two and the $f'(x)=0$ so it "flattens out" at $x = 2$.
A critical point that is neither a local maximum or minimum means (if it is differentiable) means it "flattens" out at $x=4$ but neither "plateaus" or "valleys". So it "passes through".
And as 5 is an extreme point of the interval $[1,5]$, an absolute minimum just means the lowest point on the interval. It need not be a critical point.
If you forgive the crappy graphics:
The function you've given is continuous and piecewise differentiable, but not differentiable. This means it's probably easier to look for critical points (I think you call them "intervals of monotonicity") by breaking the function down into a piecewise function of differentiable functions as follows:
$$f(x)=e^x|x-1|=e^x\left\{\begin{matrix}
x-1 & x\geq1\\
-x+1 & x<1
\end{matrix}\right.=\left\{\begin{matrix}
e^x(x-1) & x\geq1\\
e^x(1-x) & x<1
\end{matrix}\right.$$
Now we can look for critical points in each of these functions on their intervals, as well as an "automatic" critical point at the discontinuity ($x=1$). First let us examine the function for $x\geq1$:
$$f'(x)=\frac{\rm d}{{\rm d}x}\left(e^x(x-1)\right)
\\=\frac{\rm d}{{\rm d}x}(e^x)(x-1)+e^x\frac{\rm d}{{\rm d}x}(x-1)
\\=e^x(x-1)+e^x\cdot1
\\=xe^x$$
Solving this for $f'(x)=0$, we get $x=0$, which we can throw out because we are only looking at the function for $x\geq1$. Thus there are no critical points on this interval. Now we can consider the function for $x<1$:
$$f'(x)=\frac{\rm d}{{\rm d}x}\left(e^x(1-x)\right)
\\=\frac{\rm d}{{\rm d}x}(e^x)(1-x)+e^x\frac{\rm d}{{\rm d}x}(1-x)
\\=e^x(1-x)+e^x(-1)
\\=-xe^x$$
Solving this for $f'(x)=0$, we get $x=0$, which is on our interval $x<1$. We can use the second derivative test to identify this point:
$$\left.f''(x)\right|_{x<1}=\frac{\rm d}{{\rm d}x}(-xe^x)=-(x+1)e^x)$$
$$f''(0)=-(1)e^0=-1<0$$
Thus $x=0$ is a local maximum. Now we must consider the critical point at the non-differentiability. We can't really use the second derivative test to identify the point, but we can examine the first derivative before and after:
$$f'(1^+)=(1)e^1=e$$
$$f'(1^-)=(1-2)e^1=-e$$
So the slope is negative before $x=1$, and positive after. This means we have a local minimum at $x=1$
Also note that $f(1)=0$. Since $e^x$ is positive for all $x$, and $|x-1|$ is positive for all $x\neq1$, it must be the case that $f(1)$ is also the absolute minimum.
Since $e^x(x-1)$ tends to infinity as $x$ goes to infinity, there is no absolute maximum.
Best Answer
Ironically as I finished typing my question, the solution struck my mind: that $x=5$ and $x=8$ are turning points of $f$, regardless of the differentiability of $f$ at these points. Since $f'(x)>0$ for $x<5$, and $f'(x)<0$ for $x>5$, $f$, being continuous, has a local maximum at $x=5$(similar reasoning follows for $x=8$).