I have been going through a state maths exam and am not able to answer the following question. I understand that $k>-1/4$. However, in the solutions it is also stated that $e^{-x}>0$. Why does this need to happen?

Question Context


Find the values of $k$ such that the curves $y =e^{-2x}$ and $y=e^{-x}+k$ intersect at two points.

Solutions provided by exam board


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    $\begingroup$ Hint : if $~a > 0, ~$ then $~\dfrac{1}{a} > 0.~$ Is $~e^x~$ ever $~\leq 0,~$ for any $~x \in \Bbb{R}~?$ What does the graph of $~y = e^x~$ look like ? $\endgroup$ Commented Jun 3 at 4:47
  • $\begingroup$ Intuitively, note that $e^{-x}+k$ has an asymptote $y=k$. But of course this doesn't prove anything. $\endgroup$ Commented Jun 3 at 4:48
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    $\begingroup$ @BenjaminWang Perhaps I am misinterpreting the question. The way that I interpret the question, the original poster is simply asking why it is, that for all $~x \in \Bbb{R},~$ the expression $~e^{-x}~$ is always positive. $\endgroup$ Commented Jun 3 at 4:50
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    $\begingroup$ For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. $\endgroup$ Commented Jun 3 at 4:59
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    $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$
    – Martin R
    Commented Jun 3 at 5:00

1 Answer 1


So you were right to say $k>−1/4$, you say your confused with how $e^{-x} >0$,

This is simply due to the fact that $e^x$ for any real value of $x$, is positive.

Since the equations are based on $e^{-x}$, which is $1/e^x$, this would mean $1/positive$ $number$ = positive number.

Hence x is such a value below infinity, such that $e^{-x}>0$.

The reason I say below infinity is that as x -> infinity,

$1/e^x$ -> 0.

Hence the value of x is such that x is a value below infinity!

Because you cannot have a real number value such that 1/$e^x$ =0.


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