Composition and Inverse

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  • Topic: Inverse function, Function, Exponentiation
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  • Published : March 3, 2013
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Composition and Inverse

This week we have been assigned three functions which we must evaluate. These are the functions which we have to evaluate this week. fx=2x+3 gx=x2-3 hx=7-x3 We have been asked to compute(f-h)(4).

So we can evaluate each separately and then subtract.
=33=1 h4=1
This is our final answer.
Next we are to compare two pairs of the functions into each other. First we will work out. f°gx=f(gx) This means the rule of f will work on g.
=fx2-3 Here f is now going to work on the rule of g. =2x2-3+5 The rule of f is applied to g.
=2x2-6+5 Simplifying
f°gx=2x2-1 The final results.

Now we will compose the following:
h°gx=h(gx) The rule of h will work on g.
=-7+(x2-3)3 The rule of h is applied to g. h°gx=-10+x23 The final results.

Next we are asked to transform g(x) so that the graph is placed 6 units to the right and 7 units downward for where it would be right now. * Six units to the right means to put a -6 in with x to be squared. * Seven units downward means to put -7 outside of the squaring. * The new functions will look like this:

Our last job is to find the inverse of two of our functions, f and h. To find the inverse we will write the function with y instead of the function name, then we will switch the places of x and y, and solve for y again. Here are the functions: fx=2x+3 hx=7-x3

Here we replace f(x) and h(x) with y:
y=2x+3 y=7-x3
Here we switch the y and the x:
x=2y+3 x=7-y3
Now we solve for y:
Subtract 3 from both sides. Multiply both sides by 3. -3+x=2y 3x=7-y
One more solving...
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