MHB Finding Continuous Values of $\frac{e^{sinx}}{4 - \sqrt{x^2 - 9}}$

[tmt1]

I have this function

$$\frac{e^{sinx}}{4 - \sqrt{x^2 - 9}}$$

And I need to find all the values for which this function is continuous.

So I do

$$4 - \sqrt{x^2 - 9} \ne 0$$

$$ \sqrt{x^2 - 9} \ne 4 $$

$$ x^2 - 9 \ne 16 $$

$$ x^2 \ne 7 $$

And therefore, the function is not valid where

$$ x \ne +/- \sqrt{7} $$

However, this appears to be completely wrong. Why is my reasoning wrong?

 

[MarkFL]

First, I would look at:

$$x^2-9\ge0$$

This gives us:

$$(-\infty,-3]\,\cup\,[3,\infty)$$

Next, look at

$$\sqrt{x^2-9}\ne4$$

$$x^2\ne25$$

$$|x|\ne5$$

$$(-\infty,-5)\,\cup\,(-5,5)\,\cup\,(5,\infty)$$

And so, we find:

$$(-\infty,-5)\,\cup\,(-5,-3]\,\cup\,[3,5)\,\cup\,(5,\infty)$$

 

[tmt1]

First, I would look at:

$$x^2-9\ge0$$

This gives us:

$$(-\infty,-3]\,\cup\,[3,\infty)$$

Next, look at

$$\sqrt{x^2-9}\ne4$$

$$x^2\ne25$$

$$|x|\ne5$$

$$(-\infty,-5)\,\cup\,(-5,5)\,\cup\,(5,\infty)$$

And so, we find:

$$(-\infty,-5)\,\cup\,(-5,-3]\,\cup\,[3,5)\,\cup\,(5,\infty)$$

So what I had was correct but not comprehensive?

 

[MarkFL]

So what I had was correct but not comprehensive?

No, you took:

$$x^2-9\ne16$$

And added 9 to the left side while subtracting 9 from the right side. You also did not require the radicand to be non-negative. :)

 

[HallsofIvy]

I have this function

$$\frac{e^{sinx}}{4 - \sqrt{x^2 - 9}}$$

And I need to find all the values for which this function is continuous.

So I do

$$4 - \sqrt{x^2 - 9} \ne 0$$

$$ \sqrt{x^2 - 9} \ne 4 $$

$$ x^2 - 9 \ne 16 $$

$$ x^2 \ne 7 $$

No, if $x^2- 9\ne 16$ then $x^2= x^2- 9+ 9= 16+ 9= 25$, not 16- 9= 7.

And therefore, the function is not valid where

$$ x \ne +/- \sqrt{7} $$

However, this appears to be completely wrong. Why is my reasoning wrong?