Do you hail graphing calculators?

Let us say that the graphs of $y=e^x$ and $y=x$ intersect at $(a,b$ ), then certainly the two graphs must pass through $(a,b$ ).

Then $b=e^a$ and $b=a$ .

$\implies e^a - a = 0$ .

$y=e^x$ and $y=x$ don't have any discontinuous points. So if there exists any such value of $a$ , then $e^x - x$ must be equal to $0$ for some value of $a$ .

This means that if $e^x - x$ has a real root $a$ , then only there will be any intersection point.

Let us check that if $e^x - x$ can ever be equal to $0$ or not:

Let us find the maximum and minimum values of the function:

Critical points:

$\frac{d}{dx} (e^x - x) = 0$

$\implies e^x - 1 = 0 \implies x=0$

Hence $e^x - x$ is critical at $x=0$ .

Second derivative test:

$\frac{d^2}{dx^2} (e^x - x) = e^x$

Plugging in $x=0$ , the second derivative equals to $e^0 = 1$ and $1>0$ .

Hence $e^x - x$ has only a minimum value, at $x=0$

So, the minimum value of $e^x - x$ is $e^0 - 0 = 1$ .

So, the lowest value that $e^x - x$ can reach is $1$ .

This means that $e^x - x$ can never be equal to $0$ . This suggests that $e^x - x$ has no real root, hence there exists no such real $a$ for which $e^a - a = 0$ . Subsequently, there doesn't exist any such point $(a,b)$ for which the graphs of $y=e^x$ and $y=x$ intersect.

Below $x=0$ we have $y=e^x$ positive, $y=x$ negative, so they cannot be equal to each other.

At $x=0$ we have $e^0=1$ . Its derivative/slope is also $1$ and increases to the right. So $y=e^x$ is above $y=x$ , and is going up just as fast to start with and then faster. They can never intersect.