How to find domain and range of a function algebraically

1. Finding the domain of a function algebraically involves solving for the values of x that make the function undefined.

A function's domain is the set of all input values for which the function produces a result. The domain of a function can be determined algebraically by solving for the values of x that make the function undefined.

2. To do this, set the function equal to y and solve for x.

A domain is the set of all possible input values for a function. To find the domain of a function, set the function equal to y and solve for x. This will give you the set of all x-values for which the function produces a real y-value.

3. Any values of x that make the function undefined will be excluded from the domain.

A function's domain is the set of all input values for which the function produces a result. Any values of x that make the function undefined will be excluded from the domain. For example, the domain of the function f(x) = 1/x will exclude any values of x that result in a division by zero.

4. In some cases, it may be necessary to use the Quadratic Formula to solve for x.

A domain is the set of all possible input values for a function. In some cases, it may be necessary to use the Quadratic Formula to solve for x. The Quadratic Formula is a mathematical formula used to solve for the roots of a quadratic equation.

5. The domain of a function can also be graphed to visualize where the function is defined.

The domain of a function is the set of all input values for which the function produces a result. The domain of a function can also be graphed to visualize where the function is defined. The graph of the domain is a visual representation of the function's inputs and outputs.

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I've been having trouble when trying to find the domain/range of functions algebraically. Here is an example:

$P(x)=\frac{1}{3+\sqrt{x+1}}$

Finding the domain:

$x+1\ge0$

$x\ge-1$

Therefore, $x \in [-1,+\infty)$

Finding the range: Let $y=P(x)=\frac{1}{3+\sqrt{x+1}}$ From isolating x we find:

$x=(\frac{1}{y} -3)^2-1$

Therefore:

$(\frac{1}{y} -3)^2-1\ge-1$

$(\frac{1}{y} -3)^2\ge0$

$\frac{1}{y} -3\ge0$ or $\frac{1}{y} -3\le0$

$y\le \frac{1}{3}$ or $y\ge \frac{1}{3}$

This doesn't make any sense! Intuitively I can see that when $x=-1$ then $f(x)=\frac{1}{3}$ and as x approaches $+\infty$ then $f(x)$ approaches zero (without ever reaching it). How do I find this solution algebraically? What are the "rules" for working with inequalities w/ exponents and radicals (both positive and negative)? How do I find the range for other functions such as $g(x)=3+\sqrt{16-(x-3)^2}$ and $h(x)=\frac{12x-9}{6-9x}$ algebraically? A thorough explanation would be appreciated (also, feel free to point out errors in my work- there are obviously many).

NSA

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asked Jun 15, 2016 at 15:20

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You are making things more difficult than necessary in your effort to find the range. It is not really necessary to yield an inverse (as you seem to do). You could do it in simple steps:

• range of $\sqrt{1+x}$ is $[0,\infty)$
• range of $3+\sqrt{1+x}$ is $[3,\infty)$
• range of $\frac{1}{3+\sqrt{1+x}}$ is $(0,\frac13]$

answered Jun 15, 2016 at 15:40

drhabdrhab

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I don't fully follow what you are doing to determine the range. In any case, when you have:

$(1/y -3)^2\ge0$

The LHS is a square and thus always positive, this inequality is satisfied for all $y$...

In the formula: $$y = \frac{1}{3+\sqrt{x+1}}$$ the range of the monotonically increasing part $\sqrt{x+1}$ is (clearly) $[0,+\infty)$, which means the denominator is monotonically decreasing with a maximum in $x=-1$, namely $y = 1/3$. For $x \to \infty$, $y \to 0$ but since $y \ne 0$ for all $x$, the range is: $0 < y \le \tfrac{1}{3}$.

Alternatively: $$0 \le \sqrt{x+1} < +\infty$$ $$3 \le 3+\sqrt{x+1} < +\infty$$ $$ \frac{1}{3} \ge \frac{1}{3+\sqrt{x+1}} > \frac{1}{+\infty}$$ So: $$ 0 < \frac{1}{3+\sqrt{x+1}} \le \frac{1}{3}$$

For the domain of: $$g(x)=3+\sqrt{16-(x-3)^2}$$ You need: $$16-(x-3)^2 \ge 0 \iff (x-3)^2 \le 16 \iff |x-3| \le 4 \iff -1 \le x \le 7 $$ For the range (given the domain as above): $$0 \le 16-(x-3)^2 \le 16$$ $$0 \le \sqrt{16-(x-3)^2} \le 4$$ $$3 \le 3+ \sqrt{16-(x-3)^2} \le 7$$

answered Jun 15, 2016 at 15:29

StackTDStackTD

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