How do I find where (4 - 4x²)/(x² + 1)² > 0?

[mathdad]

How do I find the range of [(4 - 4x^2)/(x^2 + 1)^2] > 0 algebraically?

Do I set the numerator to 0 and solve for x?

Do I set the denominator to 0 and solve for x?

 

[mathmari]

You want that the fraction is greater than zero.

A fraction is positive either when both numerator and denominator are positive or when both are negative.

In this case the denominator is positive for each $x$. So, you have to find for which values of $x$ the numerator is also positive.

 

[mathdad]

You want that the fraction is greater than zero.

A fraction is positive either when both numerator and denominator are positive or when both are negative.

In this case the denominator is positive for each $x$. So, you have to find for which values of $x$ the numerator is also positive.

Are you saying to find the range I must set the numerator to = 0 or do I set the numerator > 0 and then solve for x?

 

[mathmari]

Are you saying to find the range I must set the numerator to = 0 or do I set the numerator > 0 and then solve for x?

We want to find the values of $x$ that satisfy the inquality $\frac{4 - 4x^2}{(x^2 + 1)^2} > 0$.

When we multiply both sides of an inequality by a negative number, the direction of the inequality changes ("<" becomes ">" for example).

In this case $(x^2 + 1)^2$ is always positive. So, multiplying the inequality by $(x^2 + 1)^2$ the direction of inequality does not change.

So, we have the following: \begin{align*}&(x^2 + 1)^2\cdot \frac{4 - 4x^2}{(x^2 + 1)^2} > (x^2 + 1)^2\cdot 0 \\ & \Rightarrow 4 - 4x^2> 0 \end{align*}

We have to solve the last inequality for $x$ to find the desired values of $x$.

 

[mathdad]

We want to find the values of $x$ that satisfy the inquality $\frac{4 - 4x^2}{(x^2 + 1)^2} > 0$.

When we multiply both sides of an inequality by a negative number, the direction of the inequality changes ("<" becomes ">" for example).

In this case $(x^2 + 1)^2$ is always positive. So, multiplying the inequality by $(x^2 + 1)^2$ the direction of inequality does not change.

So, we have the following: \begin{align*}&(x^2 + 1)^2\cdot \frac{4 - 4x^2}{(x^2 + 1)^2} > (x^2 + 1)^2\cdot 0 \\ & \Rightarrow 4 - 4x^2> 0 \end{align*}

We have to solve the last inequality for $x$ to find the desired values of $x$.

Are you saying that solving the last inequality for x will yield the range of the original inequality? I am trying to find the range.

 

[HallsofIvy]

Okay, then, what do you mean by "range" of a rational inequality? Normally, "range" refers to the set of all values of a function, but an inequality is not a function. Mathmari has told you how to solve the inequality- to find all values of x that satisfy the inequality. Is that what you mean by "range"?

 

[mathdad]

Okay, then, what do you mean by "range" of a rational inequality? Normally, "range" refers to the set of all values of a function, but an inequality is not a function. Mathmari has told you how to solve the inequality- to find all values of x that satisfy the inequality. Is that what you mean by "range"?

Again, I found this question online and posted as typed.

- - - Updated - - -

At another math site, someone responded by saying that the domain is (-infinity, infinity) and the range is [(-1/2), 4].
How is this found? Is this correct? If so, why?

 

[MarkFL]

...At another math site, someone responded by saying that the domain is (-infinity, infinity) and the range is [(-1/2), 4].
How is this found? Is this correct? If so, why?

Consider:

$$f(x)=\frac{4-4x^2}{\left(x^2+1\right)^2}$$

By looking at the above function definition, we can see that no real values of $x$ can cause $f$ to become undefined or complex, and so we may state that the domain is:

$$(-\infty,\infty)$$

To find the range, let's arrange the definition as follows:

$$fx^4+2(f+2)x^2+(f-4)=0$$

Hence:

$$x^2=\frac{-2(f+2)\pm\sqrt{4(f+2)^2-4f(f-4)}}{2f}=\frac{-(f+2)\pm2\sqrt{2f+1}}{f}$$

Now, we know:

$$0\le x^2$$

Thus:

$$0\le\frac{-(f+2)\pm2\sqrt{2f+1}}{f}$$

If we consider the case where $0<f$, then we must have:

$$f+2\le2\sqrt{2f+1}$$

$$f^2-4f\le0$$

$$f(f-4)\le0$$

So, we find:

$$0<f\le4$$

If we now consider the case where $f<0$, then we observe that the radicand cannot be negative:

$$2f+1\ge0$$

$$f\ge-\frac{1}{2}$$

And so, putting these together, we obtain:

$$-\frac{1}{2}\le f\le4$$

 

[mathdad]

Consider:

$$f(x)=\frac{4-4x^2}{\left(x^2+1\right)^2}$$

By looking at the above function definition, we can see that no real values of $x$ can cause $f$ to become undefined or complex, and so we may state that the domain is:

$$(-\infty,\infty)$$

To find the range, let's arrange the definition as follows:

$$fx^4+2(f+2)x^2+(f-4)=0$$

Hence:

$$x^2=\frac{-2(f+2)\pm\sqrt{4(f+2)^2-4f(f-4)}}{2f}=\frac{-(f+2)\pm2\sqrt{2f+1}}{f}$$

Now, we know:

$$0\le x^2$$

Thus:

$$0\le\frac{-(f+2)\pm2\sqrt{2f+1}}{f}$$

If we consider the case where $0<f$, then we must have:

$$f+2\le2\sqrt{2f+1}$$

$$f^2-4f\le0$$

$$f(f-4)\le0$$

So, we find:

$$0<f\le4$$

If we now consider the case where $f<0$, then we observe that the radicand cannot be negative:

$$2f+1\ge0$$

$$f\ge-\frac{1}{2}$$

And so, putting these together, we obtain:

$$-\frac{1}{2}\le f\le4$$

You are amazing. Thank you. Thank you also for being patient with me.

 

[HallsofIvy]

Yes, MarkFL is amazing! And do you realize that your problem, as originally stated, to find the range of [(4 - 4x^2)/(x^2 + 1)^2] > 0, is NOT the same as the problem MarkFL solved, finding range of the function, f(x)= [(4 - 4x^2)/(x^2 + 1)^2]?

 

[MarkFL]

Yes, MarkFL is amazing!...

Aw...shucks. (Blush)(Worried)