F X = 5/x 2.g X =2x 5/x Are The Functions Inverses Of Each Other

Verify Inverse Functions?

If f(x) and g(x) are inverses to each other then:
both f(g(x)) = g(f(x))= x

f(g(x)) = 5g(x) - 9
but g(x) = (x - 9)/5, so

f(g(x)) = 5[(x - 9)/5] - 9
= (x - 9) - 9 = x-18
g(f(x) = [(5x-9)-9]/5 = x-18/5
the two given functions does not seems to inverse of each other
unless there is a typo error such as g(x) = (x+9) / 5 (instead of (x-9)/5) then
in which case:
Now we have,
f(g(x)) = (5(x + 9)/5 - 9
= (x + 9) - 9
= x
g(f(x)) = [(5x - 9) + 9]/5
= 5x/5
= x

Find the inverse of the function f(x)=2x^3+5x^2+7x+7?

This is incredibly difficult to do. you are essentially trying to solve the equation
2x^3 + 5x^2 + 7x + 7 - y = 0

Unless you are lucky with the numbers involved (which you aren't here) then solving a cubic equation is very difficult. There is a formula but it is horrendous, so I won't try to copy it here. If you want to see it google "cubic equation formula".

Consider the function f(x)=(1/3)x+2?

(A) Actually, Wolfram Alpha and Mathway both give the same answer as:
3x - 6 = 3(x - 2), after factoring.

As to how f^-1(x) = 3x - 6 was derived, recall that to find the inverse of a function we need to let f(x) = y, interchange x and y, and then re-solve for y; the resulting expression in y is the inverse function.

So, here, let f(x) = y to yield:
y = (1/3)x + 2.

Now, interchanging x and y yields:
x = (1/3)y + 2.

Then, re-solving for y yields y = f^-1(x) = 3x - 6 as required.

(B) Recall that if f(x) and g(x) are inverses, then they satisfy the equality:
f[g(x)] = g[f(x)] = x.

So, we just need to check that f(x) = (1/3)x + 2 and g(x) = 3x - 6 satisfy these.

We have:
f[g(x)] = f(3x - 6), since g(x) = 3x - 6
= (1/3)(3x - 6) + 2, by replacing x with g(x) = 3x - 6
= x - 2 + 2, by distributing the 1/3
= x, by adding,

and:
g[f(x)] = g[(1/3)x + 2], since f(x) = (1/3)x + 2
= 3[(1/3)x + 2] - 6, by replacing x with f(x) = (1/3)x + 2
= x + 6 - 6, by distributing the 3
= x, by subtracting.

Since f[g(x)] = g[f(x)] = x, f(x) and g(x) are inverses as required.

(C) Notice that Wolfram Alpha's graph contains the dotted line y = x; this is to show that f(x) and g(x) are reflections of each other through y = x, a property of inverse functions. Basically, all this part of this problem wants you to do is graph f(x) and g(x) along with y = x and to be aware that you can reflect f(x) through y = x to get g(x) and vice-versa.

I hope this helps!

Inverses: f(x)=5x/3+5?

say f(x) is the variable "y"

Then y = 5x/3+5

Switch y and x so now you have
x=5y/3+5

Now solve for y again...
x=5y/3+5
x-5=5y/3 (subtract 5)
3(x-5)=5y (x3)
3(x-5)/5 = y (divide by 5)
(3x-15)/5 = y (distribute the x3)
3x/5 - 3 = y (distribute the divide 5)

This give you the inverse funtion f_inverse(x) = 3x/5-3

To check to make sure that you have done the algebra correctly, you plug in your new function into all instances of x in the original function.
y=5(3x/5-3)/3+5
You should get that y=x
y=(3x-15)/3+5 (distribute the x5)
y=x-5+5 (distribute the divide by 3)
y=x

This shows that you have the correct inverse function.

If f(x) = x^3 + 2x and g is the inverse of f, then g ' (3) is?

If f(x) = x^3 + 2x and g is the inverse of f, then g ' (3) is?

Please explain in details, thank you. This problem is really confusing me and I just need a really good break down of it. I will really appreciate it, and again thank you!