What is the inverse function of h(x)=3−x+4x−7 and how do you evaluate h−1(9)?

2 Answers
Luca F. · Stefan V.
Jun 11, 2015

h−1(y)=7y−25y−2 where y≠2.
h−1(9)=387

Explanation:

h(x)=3−x+4x−7 having x≠7
I find the common denominator and sum all together:
h(x)=3x−21x−7−x+4x−7
h(x)=2x−25x−7
Now I simplify making the division of the 2 polynomials and I obtain:
quotient=2,reminder=−11
So I can write the function as:
h(x)=2x−7x−7−11x−7
h(x)=2−11x−7.

Now, the question is how to find the inverse function? Firstly, I try to isolate x:
(x−7)(h(x)−2)=−11
(x−7)=−11h(x)−2
x=−11h(x)−2+7
Therefore we rewrite better the function:
h−1(y)=−11y−2+7=7y−14−11y−2=7y−25y−2.
So we can state that:
h−1(y)=7y−25y−2 where y≠2.

If we want to find h−1(9):
h−1(9)=7⋅9−259−2=63−257=387

Stefan V.
Jun 11, 2015

The inverse function is h−1(x)=7x−25x−2.
h−1(x)=387

Explanation:

Since the original function is not that complex, you can determine its inverse function faster by solving the function for x and switching the result for h(x).

h(x)=3−x+4x−7

h(x)=3(x−7)−(x+4)x−7=3x−21−x−4x−7

h(x)=2x−25x−7

Solve this form of the equation for x to get

hx⋅(x−7)=2x−25

x⋅h(x)−7⋅h(x)⋅7=2x−25

x⋅h(x)−2x=7⋅h(x)−25

x(h(x)−2)=7⋅h(x)−25

x=7⋅h(x)−25h(x)−2

Once you isolate x, simply switch h(x) for x to get the inverse function

h−1(x)=7x−25x−2

To evaluate h−1(9), simply substitute x with 9 to get

h−1(9)=7⋅9−259−2=387

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