History, 29.06.2019 00:00 missdee6929

How do you find the range of a function

Answers: 2

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Answer from: hannahbannana98

Let's find the domain of f( x ) = 2 / ( x - 3 )

Do you see any x guys that would cause a problem here?

What about x = 3 ?

f( 3 ) = 2 / ( 3 - 3 ) = 2 / 0 ... ouch!

So, x = 3 is a bad guy! Everyone else is OK, though.

The domain is all real numbers except 3.

What would the interval notation be?

When in doubt, graph it on a number line:

number line showing the domain is all numbers except 3

Do the interval notation in two pieces:

domain = ( -infinity , 3 ) U ( 3 , infinity )

YOUR TURN:

Find the domain of f( x ) = 5 / ( x + 7 )

Sometimes, you can't find the domain with a quick look.

Check it out:

Let's find the domain of f( x ) = 1 / ( 3 - 2x )

Hmm... It's not so obvious!

BUT, we are still looking for the same thing:

f( x ) = 1 / ( 3 - 2x ) The bad x that makes

the denominator 0!

How do we find it? Easy!

Set the denominator = 0 and solve!

3 - 2x = 0 ... subtract 3 from both sides ... -2x = -3 ... x = -3 / -2 = 3 / 2

The domain is = ( -infinity , 3 / 2 ) U ( 3 / 2 , infinity )

TRY IT:

Find the domain of f( x ) = 6 / ( 5x + 3 ) *show work!!

How about this one?

f( x ) = square root( x + 5 )

Square roots -- what do we know about square roots?

square root( 16 ) = 4 ... So, 16 is OK to put in.

square root( 0 ) = 0 ... So, 0 is OK.

square root( 3.2 ) is about 1.788 ... Yuck! But, 3.2 is OK.

square root( -25 ) = ? ... Nope! Can't do it!

*We only want real numbers!

No negatives are OK!

square root( inside )

The inside of a radical cannot be negative if we want real answers only (no i guys). So, the inside of a radical has to be 0 or a positive number.

Set inside is greater than or equal to 0 and solve it!

Now, let's find the domain of

f( x ) = square root( x + 5 ) ... x + 5 is greater than or equal to 0 ... x is greater than or equal to -5

So, the domain of f( x ) = square root( x + 5 ) is [ -5 , infinity ) .

TRY IT:

Find the domain of f( x ) = square root( 3 - x ) . *Show work!!

Here's a messier one:

Let's find the domain of f( x ) = square root( 7 - 8x )

Set

7 - 8x is greater than or equal to 0

and solve!

7 - 8x is greater than or equal to 0 ... subtract 7 from both sides ... -8x is greater than or equal to -7 ... divide both sides by -8 ... x is less than or equal to 7 / 8

The domain is ( -infinity , 7 / 8 ] .

Answer from: melissa3333

To find the excluded value in the domain of the function, equate the denominator to zero and solve for . So, the domain of the function is set of real numbers except . The range of the function is same as the domain of the inverse function. So, to find the range define the inverse of the function.

Answer from: Quest

answer

$x = \pm 2\sqrt{2} \: \: or \: \: x = \pm 1$

explanation

the given equation is

${x}^{4} - 9 {x}^{2} + 8 = 0$

we can rewrite this as:

$({x}^{2})^{2} - 9 {x}^{2} + 8 = 0$

let

$u = {x}^{2}$

then the equation becomes:

${u}^{2} - 9 {u}^{2} + 8 = 0$

the factors of 8 that sum up to -9 are {-1,-8}

we split the middle term with these factors to obtain,

${u}^{2} - u - 8u+ 8 = 0$

we factor by grouping to obtain:

$u(u - 1) - 8(u - 1) = 0$

$(u - 8)(u - 1) = 0$

apply the zero product property to get:

$u - 8 = 0 \: \: or \: \: u - 1 = 0$

$u = 8 \: or \: u = 1$

but

$u = {x}^{2}$

${x}^{2} = 8 \: or \: {x}^{2} = 1$

$x = \pm \sqrt{8} \: \: or \: \: x = \pm \: \sqrt{ 1}$

$x = \pm 2\sqrt{2} \: \: or \: \: x = \pm 1$

Answer from: Quest

Its a whole new world

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