Proving that there are 3 solutions at most $f(x)=e^x-(ax^2+bx+c)$

Question

Prove that the function $f(x)=e^x-(ax^2+bx+c)$ has 3 solutions at most .

$a$,$b$ and $c$ are constants.

This is the information given about the function, I tried a couple of things and I am not sure if what I did is right.

First the function is continuous and differentiable since $e^x$ is continuous and differentiable and $ax^2+bx+c$ is a polynomial so it is also continuous and differentiable therefore we can use rolles theorem.

I tried doing a couple of derivatives such as $f'(x)=e^x-2ax-b$ and $f''(x)=e^x-2a$ and lastly $f'''(x)=e^x$

so the third derivative has no solution , the second one is $x=ln(2a)$ since the second derivative has only 1 solution , then the first derivative has 2 at most and the original has 3 at most.

Is it the right way to solve it? am I missing something? thank you for the help! By solution I mean f(x)=0

Answer 1

The arguments you make need to be a little more precise, but the idea is correct!

Let the function $f(x)=e^x-(ax^2+bx+c)$ that is a continuous and differentiable function in $\mathbb{R}$ and let its derivative function $f'(x)=e^x-2ax-b$ that is continuous and differentiable in $\mathbb{R}$. The function $f(x)$ is in the conditions described by Rolle's Theorem on any closed and bounded interval we want.

Suppose $f(x)$ has $n$ distinct real roots with $n>3$. Let be the distinct real roots, which we call ordered from least to greatest, $x_1,x_2,x_3,\ldots ,x_n\in \mathbb{R}$. In such a case,

$$f(x_1)=f(x_2)=f(x_3)=\cdots =f(x_n)=0$$

By Rolle's Theorem, there exists at least one value in the interval $(x_1,x_2)$ where $f'(x)=0$; at least one value in the interval $(x_2,x_3)$ where $f'(x)=0$; ... at least one value in the interval $(x_{n-1},x_n)$ where $f'(x)=0$.

We have just proved the existence of at least $n-1$ roots of $f'(x)$ with $n>3$ in a correct way. That is, there exist at least $3$ roots of the equation $e^x-2ax-b=0$. Let be the distinct real roots $x_1',x_2',x_3',\ldots, x_{n-1}'\in \mathbb{R}$. In such a case,

$$f'(x_1')=f'(x_2')=\cdots =f'(x_{n-1}')=0$$

By Rolle's Theorem (applied to $f'$), there exists at least one value in the interval $(x_1,x_2)$ where $f''(x)=0$; at least one value in the interval $(x_2,x_3)$ where $f''(x)=0$,... at least one value in the interval $(x_{n-2},x_{n-1})$ where $f''(x)=0$.

We have just proved the existence of at least $n-2$ roots of $f''(x)$ with $n>3$. That is, there exist at least $2$ roots of the equation $f''(x)=e^x-2a=0$.

On the other hand, we know how to solve algebraically this equation which, as it turns out, has only one real solution other than if $a>0$:

$$e^x-2a=0\Leftrightarrow e^x=2a \Leftrightarrow x=\ln (2a)$$

And if $a\leq 0$, it doesn't exist any real solution to the equation $e^x-2a=0$ because $e^x>0, \,\, \forall x\in \mathbb{R}$.

This is an absurd thing according to what we have found above.

Therefore, the absurdity comes from considering that $f(x)$ has more than three distinct real roots. We conclude that the number of roots of $f(x)$ cannot exceed the number of three.

Answer 2

I like this question but you must correct the statement of the question as we want to solve the equation $f(x)=0$ where $f(x)=e^x-ax^2-bx-c$. And the claim is that it has at most $3$ roots or in other words $3$ solutions.

My thinking:

$f''(x)=e^{x}-2a$. We have two cases:

i) $a\leq 0$: Then $f''(x)$ is always positive. The graph of $f(x)$ is concave-up shape. So, it can cut the $x$-axis at $0$,$1$ or $2$ points. $1$ point solution is tangency situation to the $x$-axis.

ii) $a>0$: Solving $f''(x)=0$, we have an inflection point at $x=\ln(2a)$ for the graph of $y=f(x)$. Moreover, till the inflection point the first derivative $f'(x)=e^{x}-2ax-b$ is decrasing and after then increasing with only one local minumum. So, the graph of $y=f'(x)$ can cut $x$-axis at most $2$ points. That means the graph of the original function $y=f(x)$ has at most two local extrema. When it has two, one is local maximum on the left and the other is local minumum on the right side of the graph making an snake shape. With this shape, the graph can cut $x$-axis at most at $3$ points. But, in all stiuations, when $a>0$, we have at least $1$ solution.

Pavel Kocourek's answer here, gives a nice Lemma which can make the solution immediate: What is the maximum number of points at which an $n$-degree polynomial can intersect the power function $x^m, m>n$.