Animated Solution for Mathematics - Functions: If the domain of the function $f(x)=\frac{1}{\sqrt{10+3x-x^2}}+\frac{1}{\sqrt{x+|x|}}$ is $(a,b)$, then $(1+a{)}^2+b^2$ is equal to :

Welcome, future engineer! Today, we are going to embark on a journey to uncover the hidden boundaries of a function. In the world of mathematics, a function is like a machine; it only works if you feed it the right inputs.

The set of all valid inputs is what we call the domain. When we see a function like

it is not just one machine, but two machines working in tandem. For the entire system to function, both machines must be operational simultaneously. This means we are looking for the intersection of their individual domains.

The First Constraint

Let us focus on the first part, $g(x)=\frac{1}{\sqrt{10+3x-x^2}}$. We have a square root in the denominator.

This is a double-threat constraint: the value inside the square root must be non-negative to avoid imaginary numbers, and it cannot be zero to avoid division by zero. Thus, we require:

Dealing with a negative $x^2$ can be confusing, so let us multiply the entire inequality by $-1$. Remember, when you multiply an inequality by a negative number, the inequality sign flips!

We transform our condition into:

Now, we factor this quadratic. We need two numbers that multiply to $-10$ and add to $-3$. Those numbers are $-5$ and $2$.

So, we have:

Using the wavy curve method, we plot the critical points at $-2$ and $5$. The region where the product is negative is between these two points. Therefore, the domain for our first term is $x\in (-2,5)$.

The Modulus Challenge

Now, let us tackle the second term, $h(x)=\frac{1}{\sqrt{x+|x|}}$. Again, the expression inside the square root must be strictly positive: $x+|x|>0$.

This is where the modulus function tests our intuition. Let us test the regions.

If $x\le 0$, the modulus $|x|$ becomes $-x$. Our expression becomes $x-x=0$. Since $0$ is not strictly greater than $0$, this entire region is excluded.

Now, consider $x>0$. Here, $|x|=x$. Our expression becomes $x+x=2x$. Since $x>0$, $2x$ is always greater than $0$.

This is a perfect match! Thus, the domain for the second term is $x\in (0,\infty )$.

The Intersection and Final Victory

We have our two gatekeepers. The first requires $x$ to be in $(-2,5)$, and the second requires $x$ to be in $(0,\infty )$.

To find the domain of $f(x)$, we must find the overlap. Looking at the number line, the region that satisfies both conditions is $x\in (0,5)$.

This gives us our interval $(a,b)$, where $a=0$ and $b=5$. The final step is a simple calculation:

You have successfully navigated the constraints and found the answer. The final result is 26. Keep this analytical mindset, and no function will ever be able to hide its secrets from you!