Given f(x)=x^2+7 and g(x)=x-4/x. Find (g o f)(-1).
Hint: (g o f)(-1)=g(f(-1))

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- monkey*

Given f(x)=x^2+7 and g(x)=x-4/x. Find (g o f)(-1).
Hint: (g o f)(-1)=g(f(-1))

- schrodinger

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- freckles

so have you tried finding f(-1)
and then finding g(put result of f(-1) here)

- freckles

f(-1) tells us to use f(x)=x^2+7
we know to use f(x)=x^2+7 because of the f in f(-1)

- freckles

f(x)=x^2+7
we can see here the input is the thing in f( )
example:
f(a)=a^2+7
f(a+1)=(a+1)^2+7
f(input)=(input)^2+7
another example;
f(2)=2^2+7
can you find f(-1)=?

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## More answers

- monkey*

I'm a little lost, sorry.

- freckles

ok do you see that f(x)=x^2+7
and you are also to find f(-1)?

- freckles

f(x)=x^2+7
since we want to find f(-1)
we need to replace those x's with -1

- monkey*

Yes.

- freckles

so can you find f(-1)?

- freckles

|dw:1440459828779:dw|

- freckles

let me know when you have found f(-1)

- monkey*

Would it be f(-1)-1^2+7?

- freckles

well there should be an equal sign somewhere there

- freckles

|dw:1440460061483:dw|

- freckles

can you simplify (-1)^2+7?

- monkey*

8?

- freckles

yes
f(-1)=(-1)^2+7=(-1)(-1)+7=1+7=8
so f(-1)=8
so f(-1) is 8
now we had to find g(f(-1))
we just found that inside thing the f(-1) thing to be 8
replace f(-1) with 8
you are now really wanting to find g(8)

- freckles

g(8) tells us to use the expression named g

- freckles

g looks likes
g(x)=x-4/x

- freckles

just replace the old input x with the new input 8

- freckles

simplify and then you are done

- monkey*

So, g(8)=x-4/x?

- freckles

you have to replace all the x's with 8
not just one of the x's

- freckles

whatever you replace the first x with you have to replace the other x's on the other side also

- monkey*

g(8)=8-4/8?

- freckles

yes

- monkey*

So simplify 8-4/8?

- monkey*

- freckles

yes

- freckles

I don't think teachers like unsimplified answers :p

- monkey*

7.5 is my answer?

- monkey*

And no, they don't lol.

- monkey*

Thank you so much for the help!

- freckles

for me to check your answer
I have to know one this
is it \[g(x)=\frac{x-4}{x} \text{ or } g(x)=x-\frac{4}{x}\]

- monkey*

First one.

- freckles

so it is actually
g(x)=(x-4)/x not g(x)=x-4/x
so g(8)=(8-4)/8=4/8=1/2 or .5

- freckles

\[f(x)=x^2+7 \\ g(x)=\frac{x-4}{x} \\ g(f(-1)) = g((-1)^2+7)=g(1+7)=g(8)=\frac{8-4}{8}=\frac{4}{8}=\frac{1}{2}\]

- monkey*

So what would be the answer?

- freckles

well we just got that g(f(-1))=1/2
and we were looking for the value of g(f(-1))
so we are done

- monkey*

So .5 would be the answer?

- freckles

yes a half is the answer

- freckles

1/2 or .5
you choose the form
or write in the desired form your teacher wants

- freckles

I like pretty fractions so I would write 1/2 :p

- monkey*

Okay, thanks. And she wants a fraction. Could you help with one more?

- freckles

sure i can try

- monkey*

Determine the standard form of the equation of the line that passes through (-7,8) and (0,-8).

- monkey*

Determine the standard form of the equation of the line that passes through (-7,8) and (0,-8).

- freckles

point slope form for a line looks like:
\[y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line }\]

- freckles

you might want to go ahead and find m first

- freckles

if you have two points you can find m by using the following formula:
\[m=\frac{y_2-y_1}{x_2-x_1}\]

- freckles

we do have two points so we can use that formula

- monkey*

M=-8-8/0-(-7)?

- freckles

\[(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}\]
yes if you mean m=(-8-8)/(0-(-7))

- freckles

and yes you simplify this value for m that we just found

- freckles

so -8-8=-(8+8)=-16 right
and what about the bottom 0-(-7)=-(-7)=...

- monkey*

M=-16/7 ?

- freckles

bingo!
m is -16/7

- freckles

\[y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line } \\ \]
\[(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}\]
so let's go back to our point slope form for a line...
\[y-y_1=m(x-x_1) \\ \text{ we just found } m \text{ to be } \frac{-16}{7} \\ \\ \text{ replace } m \text{ with } \frac{-16}{7}\]
\[y-y_1=\frac{-16}{7}(x-x_1)\]

- freckles

now really the last thing to do is put in x1 and y1

- freckles

you can use the value I called x1 and y1 above

- monkey*

Okay, soo?

- freckles

that is all that is really left
is to replace x1 and y1
have you done that?

- monkey*

No, I'm confused about that.

- freckles

"
\[y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line } \\\]
\[(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}\]
so let's go back to our point slope form for a line...
\[y-y_1=m(x-x_1) \\ \text{ we just found } m \text{ to be } \frac{-16}{7} \\ \\ \text{ replace } m \text{ with } \frac{-16}{7}\]
\[y-y_1=\frac{-16}{7}(x-x_1)\]
"
You should see somehwere here that I called a number x1 and I called a number y1...

- freckles

remember the point (-7,8) ?

- freckles

x1 was -7
y1 was 8?

- monkey*

Yes! I feel stupid now.

- freckles

remember the last thing really is just to replace x1 with -7 and y1 with 8..

- freckles

thought this form we are looking at is point-slope form

- freckles

and your question asks for standard form

- freckles

ax+by=c <--so this is the form we want

- monkey*

So, -7x+8y=-16/7 ?

- freckles

your line should be
\[y-8=\frac{-16}{7}(x-(-7)) \\ \text{ or you could have actually used the point } (x_2,y_2) \text{ instead } \\ y-(-8)=\frac{-16}{7}(x-0) \\ \text{ doesn't matter you just have to plug in a point that is on the line for } \\ (x_1,y_1)\]
so either one we play with will give us the same answer...
you pick the first one or second one...
I honestly think the second one looks prettier

- freckles

\[y+8=\frac{-16}{7}x \\ \text{ note: all I did here was do } -(-8)=8 \text{ and } x-0=x \\ \text{ anyways we have so far } y+8=\frac{-16}{7}x \\ \text{ we want our } x \text{ terms and } y \text{ terms on the same side } \\ \text{ so subtract } y \text{ on both sides } 8=\frac{-16}{7}x-y \\ \]
in some people's eyes this may be done
by some prefer the number in front of x to be positive so you you can multiply both sides by -1
\[-8=\frac{16}{7}x+y\]
now some prefer just to also have integers coefficients and 16/7 is definitely not an integer
so you could also multiply both sides by 7
\[-56=16x+7y\]

- freckles

this is probably the most secure way to write it because it fits everyones' expectation of standard form for a line

- freckles

Ax+By=C
A is positive integer
B is integer
C is integer
I think this is the most strict definition of a line in standard form

- monkey*

I have another one just like this except the points are (9,-4) and (6,4). I do everything the same way, right?

- freckles

yep

- freckles

put in point slope form
then put in standard form

- monkey*

Okay, thank you.

- freckles

np

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