f(x)=x3−5
Replace f(x)
with y
.
y=x3−5
Interchange the variables.
x=y3−5
Solve for y
.
Since y
is on the right side of the equation, switch the sides so it is on the left side of the equation.
y3−5=x
Add 5
to both sides of the equation.
y3=5+x
Take the cube root of both sides of the equation to eliminate the exponent on the left side.
y=3√5+x
Solve for y
and replace with f−1(x)
.
Replace the y
with f−1(x)
to show the final answer.
f−1(x)=3√5+x
Set up the composite result function.
f(g(x))
Evaluate f(g(x))
by substituting in the value of g into f
.
(3√5+x)3−5
Simplify each term.
Remove parentheses around 3√5+x
.
f(3√5+x)=3√5+x3−5
Rewrite 3√5+x3
as 5+x
.
f(3√5+x)=5+x−5
Simplify by subtracting numbers.
.
Subtract 5
from 5
.
f(3√5+x)=x+0
Add x
and 0
.
f(3√5+x)=x
Since f(g(x))=x
, f−1(x)=3√5+x is the inverse of f(x)=x3−5
.
f−1(x)=3√5+x
Answer:
4/44 i think
Step-by-step explanation:
Answer:
<h2>see the answers ⤵️</h2>
Step-by-step explanation:
<h3>to understand the solving steps you need to know about</h3>
<h3>adding or subtracting with redical experience is nothing but algebraic addition or subtraction</h3><h3>let's solve:</h3>
Answer:
(h o k) (3) = 3
(k o h) (-4b) = -4b
Step-by-step explanation:
An inverse function is the opposite of a function. An easy way to find inverse functions is to treat the evaluator like another variable, then solve for the input variable in terms of the evaluator. One property of inverse functions is that when one finds the composition of inverse functions, the result is the input value, no matter the order in which one uses the functions in the combination. This is because all terms in a function and their inverse cancel each other and the result is the input. Thus, when one multiplies two functions that are inverse of each other, no matter the input, the output will always be the input value.
This holds true in this case, it is given that (h) and (k) are inverses. While one is not given the actual function, one knows that the composition of the functions (h) and (k) will result in the input variable. Therefore, even though different numbers are being evaluated in the composition, the output will always be the input.
