Vtgyjfy

My question is specifically addressed to the people who have experience in grading students, however everybody's opinion is very much welcomed! The more, the better. Thank you. I do understand that every teacher might have different grading method so there's no the only one correct way to go.

I had this simple problem on the exam. My answer was wrong due to very minor calculating mistake, however the way I solved my problem was correct. I would like to know how would YOU grade my submission out of total 5 points to gain? :)

Please note: I will use just a little bit shortened notation here and there, however not in crucial grading points.

Task:

Check the monotonicity and find local extreme values of $f(x) = (x^2+2x)e^{-x}$

Domain of $f$:

$e^{-x} = frac{1}{e^x} Rightarrow e^x neq 0$

Conclusion: $e^x$ is always different than $0$, because it's always positive (exponentation never equals 0), thus:

$Rightarrow D_{f} = mathbb{R}$

$f'(x) = (2x+2)e^{-x} + (x^2+2x)e^{-x} (-1)$

$f'(x) = (2x+2)e^{-x} - (x^2+2x)e^{-x}$

//here same conclusion about $D_{f'}$ // $Rightarrow D_{f'} = mathbb{R} = D_{f}$

$f'(x) = (2x+2)e^{-x} - (x^2+2x)e^{-x}$

$f'(x) = 0$

$Rightarrow (2x+2)e^{-x} - (x^2+2x)e^{-x} = 0 quad quad / :e^{-x}$ $quad$($e^{-x}$ is always positive)

$(2x+2) - (x^2+2x) = 0$

$-x^2 = -2$

$x^2 = 2$

$x = 2 quad lor quad x = -2 quad quad$ here is the terrible fail

So at this point I made a mistake. Anyway, continuing:

--At this point I nicely drew the line showing monotonicity and extremas, writing down the intervals where it's increasing/decreasing. It's just I took the $2, -2$ arguments instead of $sqrt{2}, - sqrt{2}$-- I am not sure how to draw the line etc in LaTeX so let's skip that and assume it's 100% correct. Continuing...

$f_{max}(2) = (2^2+2*2)e^{-2} = 8e^{-2} = frac{8}{e^{2}}$

$f_{min}(-2) = ((-2)^2 + 2*(-2))e^{-2} = 0e^{-2} = 0$

Full answer: (such form of an answer is desired by the teacher) Function $f$ has a local minimum at $x = -2$ equal to $0$ and a local maximum at $x=2$ equal to $frac{8}{e^{2}}$. The function is decreasing on the following intervals: $(-infty; -2), (2, infty)$ and increasing on the interval $(-2, 2)$.

TLDR: how would you rate me out of 5 points to gain? We can assume that the only mistake I did was the calculating error $x^2 = 2 Rightarrow x = 2 lor x = -2$. However, because of that mistake, I have got two wrong extremas: should be $x_{1} = sqrt{2}, x_{2} = -sqrt{2}$, and thus I also got wrong values of $f_{max}({x_{1}})$ and $f_{min}({x_{2}})$.

Feedback, opinions are appreciated! Thanks for taking your time.

There doesn't seem to be much to do about the grade, but you could do lots more for yourself: order things up! Don't write the derivative of the function the way you did, in particular if you'll use it for something else later. Write instead

$$f'(x)=(2x+2)e^{-x}-(x^2+2x)e^{-x}=e^{-x}left(2-x^2right)$$

Doing a very little algebra you can make things way clearer for you and for the grader, and now:

$$f'(x)=0iff 2-x^2=0iff x=pmsqrt 2$$

Now, the mistake you did was not because you wrote something wrong before the fatal mistake. Yet not having things nicely exposed before your eyes when you were writing down your exam may have decisively affected (in fact, impaired) a little your ability to see little things....enough to commit a costly mistake.

I, teaching now engineering students, would probably take 1 or may 2 points out of 5 possible. Yet this can go up to 3 or even four points less if the lecturer stressed a lot the ways one has to avoid this kind of mistake.

For example, a simple little thing: what about the sign of the derivative? When having enough time, I'd rather have students to check that, or even going to the second derivative (which is pretty simple in this case).

And how in the world did you arrive to the conclusion that $;2;$ is a maximum and $;-2;$ is a minimum?! This alone would cost you for not explaining things. In any case, it will be reasonably easy to spot your mistake (the sign of the first derivative would lead you immediately to deduce the derivative doesn't vanish at $;x=2;$ , for example!)