Given that f(x) = 2x + 1 and g(x) = −5x + 2, solve for f(g(x)) when x = 3.
Okay. Do you know the properties/ how to plug these in accordingly?
no i dont understand
Okay I am going to try to explain this as well as i can. f(g(x)) refers to you taking g(x) and plugging it into x for f(x) g(f(x)) refers to you taking the value of f(x) and plugging it into x in the g(x) value.
So looking at your problem you are trying to solve for f(g(x)) which means f(x) = 2 (-5x+2) Realize that they give you your x value. Which is 3. Can you plug this in now and see what you get?
bu where does the 3 go?
It goes in for the x value youve just created. When you do f(g(x)) or g(f(x)) you replace the current x value from the originall f or g equation. So, with this being said. You plug in the value according to how they want it and then you simplify. The value of x is 3. f(x) = 2 (-5(3) + 2) can you simplify it now?
oh that makes more sense. I thought it would be something like f(2x + 1(−5x + 2(3)) and i was so confused. so f(x) = 2 (-5(3) + 2) would be f(x) = -26?
but the options are −91 −33 −25 35
Okay so, my steps were f(g(x)) = 2(-5x + 2) +1 f(g(x)) = 2(-5(3) + 2)) + 1 f(g(x)) = 2 (-15+2) +1 Now order of operations. f(g(x)) = 2(-13) +1 f(g(x)) = -26 + 1 f(g(x)) = -25
Do you see how I got this answer?
yes that makes so much sense now!! thank you so much!
do you think you could help me with some more...?
Yeah sure thing just post them (:
how do i find the inverse of f(x) = x2 − 9?
f(x) = 2x + 9 let y = 2x + 9 interchange x and y x = 2y + 9 now make y the subject of the formula x = 2y + 9 x - 9 = 2y (x - 9)/2 = y f‾¹(x) = (x - 9)/2
Do you need more of an explanation?
Also if youre seeing boxes refresh the page and they go away (:
oooo wait I read the problem wrong!!!
so you are pretty much switching x for y and solving for y?
these are the options
I am so sorry :(! Here this is better for the explanation . f(x) = x^2-9 y=x^2-9 x= y^2 -9 +9 +9 x+9= y^2 Now take the square root of the question and you should get sqrt.(x+9) = y
+/- by the way. forgot that (:
so answer a is correct
ooooh thank you!
No problem (: Any more you need help with?
yes!
ask away haha (:
Given the function f(x) = 2x - 5/3, which of the below expressions is correct ^^^thats 2x - 5 divided by 3
Like this you do, we have been given f(x)=2x-5/3 so you can take that as also, y=2x-5/3 shift the 3 to the other side and division changes to multiplication, 3y=2x-5 now shift 5 to the other side and -ve becomes +ve, 3y+5=2x, divide both sides by 2, so you get, 3y+5/2=x now you can also write x as f-1(y)=3y+5/2 I just made f(x) as y to make it easier, of course there are many ways to do it At the last, you can change it back to f-1(x) and you will get the same ans as: 3x+5/2 or answer C
why does the f(x) gets changed into f^-1(x)?
Because youre essentially looking for the other form which is your inverse for (f^-1)(x)
so for this problem i switch x and y, solve for y and when i get the answer i change y by f^-1(x)?
wait. not solve. i meant get y alone
Well you could do it that way yes. BUT you could also do (this may be easier for you.) f(x) = 2x-5/3 y = 2x-5/3 Multiply both sides by 3 to get rid of the denominator. 3y=2x-5 add 5 to both sides. 3y+5 = 2x divide by 2 (3y+5)/2 = x Now you can do your inverse property and simply switch x and y y= (3x+5)/2
oh now i get it! i just did another problem like this and got the right answer!
AWESOME! You seem to be catching on (:
i have like 2 or 3 more problems that i need some help on..
Find the average rate of change for the given function from x = −1 to x = 2.
Im stuck i kinda dont remmeber this one lol. I may if I look up some of it. do you have any formulas I could maybe go off of?
one sec
okay (:
this is what the textbook gives me
Okay sorry on the wait one second (:
its ok :)
Okay so by looking at the graph you are meant to find your y coordinates that correspond to the x coordinates. can you tell me those (im walking you through step by step) (:
so would a y coordinate be 1?
uhm no. one of the coordinates would be -1 because you should look at when you see the line pass through either underneath. above or through the x coordinate.
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