How many roots does this exponential equation have?

Question

I have the following course assignment: Calculate the number of roots of the following equation: $ e^x = 6 \cdot \ln(x^2 + 1) $

What I have done so far:

$ f(x) = e^x - 6 \cdot \ln(x^2 + 1) $

I will look for the roots of the derivative and check where the function crosses OX as the function is continuous (Darboux): $ f'(x) = e^x - \frac{12x}{x^2+1} $

$f'(x) \gt 0, x \leq 0 $

$f'(1) \lt 0$

$f'(2) \gt 0$

$ f(0) = 1, f(1) \lt 0, f(2) \lt 0 $

And there's the function graph:

I am missing the part (0,1) . Can somebody give me a hint? I guess it will look as I have drawn, but I cannot prove it.

Answer 1

Since $f'(x)=\frac{e^x(x^2+1)-12x}{x^2+1}$, setting $g(x)=e^x(x^2+1)-12x$ gives us $$g'(x)=e^x(x+1)^2-12,\ \ g''(x)=e^x(x+1)(x+3).$$ Since $g'(-3)\lt 0,g'(1)\lt 0,g'(2)\gt 0$, we can see that there is only one real number $\alpha\ (1\lt \alpha\lt 2)$ such that $$g'(x)\lt 0\ (x\lt \alpha),\ g'(\alpha)=0,\ g'(x)\gt 0\ (x\gt\alpha).$$ So, since $g(0)\gt 0,g(1)\lt 0,g(2)\gt 0$, we can see that there are only two real numbers $\beta,\gamma\ (0\lt\beta\lt 1\lt\gamma\lt 2)$ such that $$g(\beta)=g(\gamma)=0,$$ which means $f'(\beta)=f'(\gamma)=0$. From what you've got, we know that the number of roots is three.

Answer 2

You have most of the steps so far. I'll finish the problem off for you.

First, let's define our function $f(x)$ as:

$$f(x) = e^x - 6\ln(x^2+1)$$

Now, we can use the Intermediate Value Theorem to find our points that become the roots of this equation. We need to take the derivative in order to find them:

$$f'(x) = e^x - \frac{12x}{x^2 + 1}$$

Now, we set equal to $0$ to find our x values:

$$e^x - \frac{12x}{x^2 + 1} = 0$$

We could find the values algebraically, but consider this:

If $x < 0$, it follows that both our fraction and our exponential term will be positive. Therefore, $f'(x) > 0$ for all $x < 0$.

Next, we consider when $x > 0$. Now, initially the fraction term will "conquer" and become a greater value than the exponential term for small values of $x$. However, as $x$ increases, the exponential term $e^x$ will ultimately increase at a higher rate over the fraction term.

Therefore, we have $2$ positive values of $x$ where $f'(x) = 0$.

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We can also say that $f'(x)$ changes from +, to -, and then +. This corresponds to $f(x)$ initially increasing, then decreasing, and then increasing again.

Now, in order to determine our roots. Let's see if we can find some trivial points.

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For instance, if $x= 0$, our function f is $e^0 - 6\ln(1) = 1$

Again, if $x= -1$, our function f is $e^{-1} - 6\ln(2) < 0$, since $e^{-1}$ is a very small term compared to $6\ln 2$.

Since $f(x)$ is increasing on this interval, by the IVT we have one root in that range of $x$ values.

Next, let's try $x = 1$. $f(x)$ would then be $e^1 - 6\ln(2)$. $\ln 2$ is close to 2, so $6\ln 2$ compared to $e$ would be much greater and therefore $f(x) < 0$ in this case.

Therefore, by the IVT we have another root in that range of $x$ values.

Finally, we also mentioned that for larger values of $x$, the derivative $f'(x)$ would eventually increase faster than the rational term, so $f'(x) > 0$.

Therefore, $f(x)$ will eventually go through the x-axis again, so we have one more zero by IVT.

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Therefore, we have $3$ zeros in total, looking at it analytically rather than algebraically.

Answer 3

look at $$f(x) = e^x - 6\ln (1 + x^2) = 1 + x + \cdots $$ therefore note that $f(0) = 1, f'(0) = 1.$

now, $$f'(x) = e^x - 12x/(1 + x^2) $$ has two zeros one in $(0,1)$ and another in (1,2). we show that the second one is local min and the function value is negative because $f(1)< 0$ and $f(2) < 0$. that should show the existence of a root in $(0,1).$ the other root in $(2,\infty)$ follows from the fact $f(\infty) = \infty$