Question

These last three problems are really stumping me, could someone please help explain them. I can't figure out where I'm messing up at.

1. Given f(x) = 2x − 3, find the following:
a) f(y) = ?

2. Given g(x) = 5x^2 − x + 2, find the following.
a) g(x + h) = ?

3. Given h(r) = 1/r + 5, find the following.
a) h(x2) + 1 = ?

Answer 1

h(x^2)+1

sorry I typo for number 3.

Answer 2

f(y)=2y-3

Answer 3

whatever is in the (), replace the variable with

Answer 4

f(spongebob) = 2(spongebob) - 3

Answer 5

ah the old
\[f(x+h)\] so beloved by math teachers.

Answer 6

f(apple pie) = 2(apple pie) - 3

Answer 7

f(g(x)) = 2(g(x)) - 3

Answer 8

\[f(\spadesuit+\heartsuit)\]

Answer 9

f(♫♫♫) = 2(♫♫♫) - 3

Answer 10

beat me to it lol

Answer 11

\[g(x) = 5x^2 − x + 2, \]
\[g(\spadesuit+\heartsuit)=5(\spadesuit+\heartsuit)^2-(\spadesuit+\heartsuit)+2\]

Answer 12

replace
\[\spadesuit\] by
\[x\] and
\[\heartsuit\] by \[h\] and you will have it

Answer 13

the notation simply tells you the name of a function, and what its value depends upon

Answer 14

Good grief, for number 1 I was trying to solve for y and then plug it in for x or something like that . . . gah I feel so dumb now.

For the second problem, I keep ending up with 5x^2+10xh+5h^2-x+h+2, but that's not the right answer.

And for the third one, I thought the answer would be 1/x^2+6, but thats not right either . . . uughh

Answer 15

f is a generic name; and x is the generic variable that the function depends upon for its value

Answer 16

the second one can be tricky since the algebra is more complicated, needs to keep an eye on it

Answer 17

Kk, let me redo number two again, my math has to be off somewhere. >_<

Answer 18

\[g(x) = 5x^2 − x + 2\]
\[ g(x+h) = 5(x+h)^2 − (x+h) + 2\]
\[ = 5(x^2+2xh+h^2) − x-h + 2\]
\[ = 5x^2+10xh+5h^2 − x-h + 2\]
is what i get

Answer 19

For \(g(x) = 5x^2 − x + 2\), \(g(x+h)\) tells to replace the variable \(x\) with \(x+h\). That results in \(g(x+h)=5(x+h)^2-(x+h)+2\). Simplify from there:

\(5(x^2+2hx+h^2)-x-h+2\)
\(5x^2+10hx+5h^2-x-h+2\)

Resembling anything like that answer you are supposed to get yet?

Answer 20

your (-x+h) part is off; you have -x+h

Answer 21

typoed the correction lol

Answer 22

Just to get over with the symbolic difficulties, I tend to use apples and cows with my students before moving to numbers. :P And doing maths without numbers altogether is a joy.

Answer 23

omg the -h!!!!! I was keeping it positive!! I forgot to carry the - over >_<

Answer 24

\[-(x+h)=-x-h\]
you have\[-(x+h)=-x+h\]

Answer 25

thats right :)

Answer 26

is the format off on your monitors as well?

i typed this on 2 lines and it comes across as one on mine

Answer 27

i might\[\]
have to type\[\]
in some delimiter

Answer 28

okay now I just have to figure out where I'm going wrong with number three and the nightmare is over . . .

I worked it out like this the first time:

h(x^2)+1 = 1/((x^2) +1) +5

Answer 29

And then ended up with 1/X^2+6 as my answer.

Answer 30

it it this ?\[h(r) = \frac{1}{r} + 5\]

Answer 31

if so then
\[h(x^2) + 1=\left(\frac{1}{r^2}+5\right)+1\]

Answer 32

umm no, on my paper it has h(r) = 1/r+5; and it's all under the /

Answer 33

ok, then just slip the +5 back under :)

Answer 34

same concept tho

Answer 35

and to fix my typos:

\[h(x^2)+1=\left(\frac{1}{x^2+5}\right)+1\]

Answer 36

oh wait O_O so I don't add the +1 under the fraction, it stays out to the side because its not included in the parentheses?!

Answer 37

its not a part of h(r) so it all on its own; use what they DEFINE h(r) to be, and dont deviate from that

Answer 38

so then it would be (1/x^2+5) +1

Answer 39

yes

Answer 40

omg why do I make this so complicated >_< It makes so much sense now, but a min ago I was completely stumped and had no clue where I was going wrong with my answers.

Thank you so much for explaining all of that, I very much appreciate it.

Answer 41

youre welcome :)