The radicand within a square root symbol must be nonnegative, so

$\displaystyle 25 - x^{2} \geq 0$

$\displaystyle 25 \geq x^{2}$

$\displaystyle x^{2} \leq 25$

This happens if and only if $\displaystyle -5 \leq x \leq 5$, so the domain of $\displaystyle f$ is $\displaystyle [-5,5]$.

$\displaystyle f (x) = \sqrt{25-x^{2}}$ assumes its greatest value when $\displaystyle 25-x^{2}$, which is the point on $\displaystyle [-5,5]$ where $\displaystyle x^{2}$ is least - this is at $\displaystyle x = 0$.

$\displaystyle f (0) = \sqrt{25-0^{2}} = \sqrt{25} = 5$

Similarly, $\displaystyle f (x) = \sqrt{25-x^{2}}$ assumes its least value when $\displaystyle 25-x^{2}$, which is the point on $\displaystyle [-5,5]$ where $\displaystyle x^{2}$ is greatest - this is at $\displaystyle x = \pm 5$.

$\displaystyle f (5) = \sqrt{25-5^{2}} = \sqrt{0} = 0$

$\displaystyle f (-5) = \sqrt{25-\left (-5 \right )^{2}} = \sqrt{0} = 0$

Therefore, the range of $\displaystyle f$ is $\displaystyle [0, 5]$.

The radicand within a square root symbol must be nonnegative, so

$\displaystyle 16 - x^{2} \geq 0$

$\displaystyle 16 \geq x^{2}$

$\displaystyle x^{2} \leq 16$

This happens if and only if $\displaystyle -4 \leq x \leq 4$, so the domain of $\displaystyle f$ is $\displaystyle [-4,4]$.

The natural domain of the composite function $\displaystyle f \circ g$ is defined to be the intersection two sets.

One set is the natural domain of $\displaystyle g$. Since $\displaystyle g$ is a polynomial, its domain is the set of all real numbers.

The other set is the set of all values of $\displaystyle x$ such that that $\displaystyle g(x)$ is in the domain of $\displaystyle f$. Since the radicand of the square root in $\displaystyle f$ must be nonnegative, 

$\displaystyle x- 4 \geq 0$, and $\displaystyle x \geq 4$, the domain of $\displaystyle f$

Therefore, the other set is the set of all $\displaystyle x$ such that

 $\displaystyle g(x) \geq 4$

Substitute:

$\displaystyle x^{2}+8 \geq 4$

$\displaystyle x^{2} \geq -4$

This holds for all real numbers, so this set is also the set of all real numbers.

The natural domain of $\displaystyle f \circ g$ is the set of all real numbers.

The domain includes the values that go into a function (the x-values) and the range are the values that come out (the $\displaystyle f(x)$ or y-values). A sine curve represent a wave the repeats at a regular frequency. Based upon this graph, the maximum $\displaystyle f(x)$ is equal to 1, while the minimum is equal to –1. The x-values span all real numbers, as there is no limit to the input fo a sine function. The domain of the function is all real numbers and the range is $\displaystyle -1\leq f(x)\leq1$.

A function has to pass the vertical line test, which means that a vertical line can only cross the function one time.  To put it another way, for any given value of $\displaystyle x$, there can only be one value of $\displaystyle y$.  For the function $\displaystyle x=\left | y\right |$, there is one $\displaystyle x$ value for two possible $\displaystyle y$ values.  For instance, if $\displaystyle y = 2$, then $\displaystyle x = 2$.  But if $\displaystyle y = -2$, $\displaystyle x = 2$ as well.  This function fails the vertical line test.  The other functions listed are a line,$\displaystyle 6x - 3y = 11$, the top half of a right facing parabola, $\displaystyle y = \sqrt{x-9} +3$, a cubic equation, $\displaystyle y = x^{3} -x +5$, and a semicircle, $\displaystyle y = -\sqrt{x^{2}-16}$. These will all pass the vertical line test.

The domain is the set of possible value for the $\displaystyle x$ variable. We can find the impossible values of $\displaystyle x$ by setting the denominator of the fractional function equal to zero, as this would yield an impossible equation.

$\displaystyle \sqrt{x^2+1}=0$

Now we can solve for $\displaystyle x$.

$\displaystyle x^2+1=0$

$\displaystyle x^2=-1$

There is no real value of $\displaystyle x$ that will fit this equation; any real value squared will be a positive number.

The radicand is always positive, and $\displaystyle f(x) = \frac{1}{\sqrt{x^{2}+1}}$ is defined for all real values of $\displaystyle x$. This makes the domain of $\displaystyle f$ the set of all real numbers.

To find the domain, find all areas of the number line where the fraction is defined.

$\displaystyle x^2-2x+1\neq0$ because the denominator of a fraction must be nonzero.

Factor by finding two numbers that sum to -2 and multiply to 1.  These numbers are -1 and -1.

$\displaystyle (x-1)(x-1)\neq0$

$\displaystyle x\neq 1$

Using $\displaystyle -1$ as the input ($\displaystyle x$) value for this equation generates an output ($\displaystyle y$) value that contradicts the stated condition of $\displaystyle y > 0$.

Therefore $\displaystyle -1$ is not a valid value for $\displaystyle x$ and not in the equation's domain:

$\displaystyle y = (-1)^2 + \frac{7}{(-1)} \rightarrow y = 1 - 7 \rightarrow y= -6$

This function is a parabola that has been shifted up five units. The standard parabola has a range that goes from 0 (inclusive) to positive infinity. If the vertex has been moved up by 5, this means that its minimum has been shifted up by five. The first term is inclusive, which means you need a "[" for the beginning.

Minimum: 5 inclusive, maximum: infinity

Range: $\displaystyle [5, \infty)$

The domain represents the acceptable $\displaystyle x$ values for this function. Based on the members of the function, the only limit that you have is the non-allowance of a negative number (because of the square root). The square and the linear terms are fine with any numbers. You cannot have any negative values, otherwise the square root will not be a real number.

Minimum: 0 inclusive, maximum: infinity

Domain: $\displaystyle [0,\infty)$