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If the tangent drawn to the curve $f(x)=-e^{-x}(x^2+2x+2)+x$ at $(a,f(a))$ meets the curve again at two distinct points and the range of $a$ is $(m,n)$. Find $m+n$

It is possible to follow the general method, ie. by finding slope at point $(a,f(a))$, then finding equation of tangent and then solving it with the given curve by discriminant $>0$

But given the relative complexity of the function, it seems that some alternative method is to be used. However I am not able to figure it out

stoic-santiago
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Aditya
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1 Answers1

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

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

So we can observe that $f''(x)>0$ on the interval $x\in(0,2)$ and $f''(x)<0$ for $x\in(-\infty,0)$ and $x\in(2,\infty)$.

Thus the curve $y=f(x)$ has points of inflection at $x=0$ and $x=2$.

So the point $a$ should lie on the interval $(0,2)$ if the said condition is to happen. So $m+n=2$

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The above is an approximate situation. I will try to be more rigorous and rule out the other cases

Maverick
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  • I didn’t get the point of inflection logic – Aditya May 16 '21 at 13:15
  • In continuous curves if sign of double derivative changes across a point then that point is point of inflection. What is the answer given – Maverick May 16 '21 at 13:16
  • two is correct . – Aditya May 16 '21 at 13:47
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    I don't think this is quite right. There are domains neighboring $(0,2)$ for which the tangent at $(a,f(a))$ still intersects twice the curve elsewhere (take $a=-0.4$ for instance, the tangent at this particular $a$ intersects the curve around $(1.654,0.115)$ and $(7.983,7.955)$). The answer $(0,2)$ works only if you additionally require $(a,f(a))$ to be "between" the two intersection points. – user3733558 May 17 '21 at 11:30
  • True. It appears that some additional information may be required – Maverick May 17 '21 at 11:33