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f(x) = (√x)/3 , then f ^-1 (x) =
We have the function f(x) = sqrtx/ 3
We need to find the inverse f^-1 (x)
==>let y = sqrtx / 3
Let us multiply by 3:
==> 3y = sqrtx
Now we will square both sides:
==> (3y)^2 = x
==> x= 9y^2
Now we will rewrite x and y and y as x:
==> y = 9x^2
Then, the inverse of the function f(x) is:
f^-1 (x) = 9x^2
We have the function f(x) = ( sqrt x)/3.
Now we need to find the iverse of teh function f(x) = (sqrt x)/3.
Let f(x) = (sqrt x)/3 = y
square both the sides
=> x / 9 = y^2
=> x = 9y^2
Replace x with y and vice versa.
We get the inverse function of f(x) = ( sqrt x)/3 as f(x) = 9x^2.
f(x) =( sqrtx)/3.
To find the f^-1(x).
Let f^-1 (x) = be y.
Then by defintion. x = f(y).
Therefore by definition if f unction f(x) = (sqrtx)/3,
f(y) = (sqrty)/3.
x = (sqrty)/3
Now we solve for y by squaring both sides:
x^2 = y/ 9
Therefore y = x^2/9 .
Therefore the inverse of f(x) = (sqrtx)/3 is y = x^2/9.
Each time when we want to determine the inverse function, we'l have to writ x with respect to y.
We'll note f(x) = y and we'll re-write:
y = (√x)/3
We'll multiply both sides by 3:
3y = (√x)
We'll raise to square both sides, to get rid off the square root:
(3y)^2 = x
We'll use the symmetric property:
x = 9y^2
So, f^-1(x) = 9x^2
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