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1-36.

Consider the functions $f(x)=3x^2−5$ and $g(x)=\sqrt{x-5}+2$

1. Find $f(5)$.

Substitute $5$ for every $x$. ( $f(x)$)

$f(5)=75−5$

$f(5)=70$

1. Find $g(5)$.

Substitute $5$ for every $x$. ( $g(x)$)

$g(5)=\sqrt{0}+2$

1. Find $f(4)$.

Substitute $4$ for every $x$. ( $f(x)$)

$f(4)=48−5$

$f(4)=43$

1. Find $g(4)$.

Substitute $4$ for every $x$. ( $g(x)$)

$g(4)=\sqrt{-1}+2$

1. Find $f(x)+g(x)$.

$f(x)+g(x)=3x^2+\sqrt{x-5}-3$

1. Find $g(x)−f(x)$.

Subtract the first equation from the second.

1. Describe the domain of $f(x)$.

Notice that the exponent is on the $x$, not the $f(x)$.

The domain of $f(x)$ is all real numbers.

1. Describe the domain of $g(x)$.

Notice that the square root sign is over the $x$.

The domain of $g(x)$ is all numbers greater than or equal to $5$.

1. Why is the domain of one of these functions more restrictive than the other?

See hints for parts (g) and (h).

They are different because the square root of a negative is undefined, whereas any real number can be squared.

Use the graphed functions in the eTool below to answer each part.
Click the link at right for the full version of the eTool:  1-36 HW eTool