Given the function f(x)=4(x+3)-5, solve for the inverse function when x=3.
plz help
@Mehek14
@Michele_Laino
To find the inverse of a function, you always want to change x and y, then solve for y again. First, change f(x) to y to avoid any confusion. \[y=4(x+3)-5\] Now, swap x and y. \[x=4(y+3)-5\] All you have to do now is solve for y. Once you've done that, substitute 3 in for x, and you'll be all set!
Wait so how do you substitute x for 3?
@adrianna45
@nincompoop can you please help?
@perl
by plug in x=3
Just replace x to 3.
so it would be x=4(y+3)-5 x=4(3+3)-5?
@geerky42
Are you still working on this ?
yes
f(x) = y = 4(x+3) - 5 is a line that is given y = 4x + 12 -5 y = 4x + 7
ya so when do you substitute x for 3?
The inverse of that , will be another line that is reflected over the line with slope 1, y=x. To figure that , you can just switch x and y around and resolve for the new y.
Original: y = 4x + 7 Inverse: x = 4y + 7 x - 7 = 4y y = (x-7) / 4
Put x=3 in the inverse function, calculate the y value
-1?
That's correct.
See it is a reflection if you switch the x and y around, as in the graph
ya i see it. it really helped thank you so much and do you know how to do f(g(x) problems?
sure
well i need help with Given that f(x) = x2 + 2x + 3 and g(x) = quantity of x plus four, over three, solve for f(g(x)) when x = 2.
like i dont know how to setup
solve g(2) first, what is that
g(x) = (x+4) / 3
2?
wait do you put 2 for x?
yes, now find f[g(2)], what do ya think to do?
4x2x+3 than 6x=3 which is 2?
since g(2) = 2 , start from the middle and work outwards f(2) = 2^2 + 2*2 + 3
11?
yeah
oh i forgot to substitute x with 2 for 2x
so if i haveanother problem like in future do i solve g first? or depending on the problem
for compound functions like f([g(x)] start with the inner function, evaluate if given a value, then use g(x) value in f(x)... work inner function to outer functions
ohok thank you very much!
welcome, goodluck
there are like some problems like f(x) and g(x) and you have to compare them. there are lot of problems but i dont know how to compare like if you know how to do it i will put it on new page
sure, just tag my name with @name
ok thanks
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