If f(x) = 5x, what is f^-1(x)?
Need help pls
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Answer:
f^-1(x) =1/5x
Step-by-step explanation:
For Inverse graphs, you interchange the x-values and the y-values.
Y=5x
Inverse:
x=5y (interchange x and y)
Change the values again, you'll get something like this -5y=-x
Divide both sides of the equation by -5
You'll get y = -x/-5
Using laws of exponents x/5 can be written as 1/5x
So y=1/5x
[tex]f^{-1}(x) = \frac{1}{5}x\\[/tex]
Given:
[tex]f(x) = 5x[/tex]
Let [tex]y = f(x)[/tex] so we'll have the equation, [tex]y = 5x[/tex]. We then write it in its inverse form where we switch the [tex]x[/tex] and [tex]y[/tex]. The equation will be [tex]x = 5y[/tex]. Now we can solve for inverse [tex]y[/tex].
[tex]x = 5y \\ 5y = x \\ 5y \times \frac{1}{5} = x \times \frac{1}{5} \\ y = \frac{1}{5}x[/tex]
Since [tex]y[/tex] is [tex]f(x)[/tex], the inverse of [tex]y[/tex] should be the inverse of [tex]f(x)[/tex] or it is [tex]f^{-1}(x)[/tex]
We can then substitute inverse [tex]y[/tex] with [tex]f^{-1}(x)[/tex]