I have this function: $$g(t)=e^{-4(t-10)}(1-4(t-10)) 1(t-10)$$ where $1(t-10)$ is unit step function.

I need to find extreme point. On WolframAlpha solution is $t=10.5$

Also I have solution to this problem in my book but I don't understand it. The solution is: $$g'(t)=(16t-168)e^{-4(t-10)}=0$$ $$t=10.5$$ I don't understand what happened to unit step function when we found derivative. It would make sense if $e^{-4(t-10)}(1-4(t-10)) =0 $ for $t=10$ but that's not the case. So I thought that derivative of this function is: $$g'(t)=(16t-168)e^{-4(t-10)}+e^{-4(t-10)}(1-4(t-10))\delta(t-10)$$

where $\delta(t-10)$ is Dirac fuction.

Can you please help me solve this. Am I missing something?


The function is identically $0$ when $t<10$, so you only need to analyze the case when $t\ge 0$. You can find the extreme on $t>10$ by finding a critical point. In general, you should compare the extreme value so obtained to $f(10)$, but that cannot be an extremum, since it is the average of the left- and right-hand limits.

The Dirac $\delta$ function doesn't come into a problem like this. The old-fashioned things you learned in calculus still work.

EDIT Perhaps one more remark is in order. You can see that $\lim_{x\to 10+}=1,$ but there is no point at which the function actually attains the value $1$. The function has no maximum.


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