How do calculate net pay and gross pay together to make one whole sum?
1 answer:
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Answer:
Step-by-step explanation:
we know that
To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.
we will proceed to verify each case to determine the solution of the problem
<u>case A)</u>
Find the inverse of f(x)
Let
y=f(x)
Exchange variables x for y and y for x
Isolate the variable y
Let
therefore
f(x) and g(x) are inverse functions
<u>case B)</u>
Find the inverse of f(x)
Let
y=f(x)
Exchange variables x for y and y for x
Isolate the variable y
Let
therefore
f(x) and g(x) are inverse functions
<u>case C)</u>
Find the inverse of f(x)
Let
y=f(x)
Exchange variables x for y and y for x
Isolate the variable y
fifth root both members
Let
therefore
f(x) and g(x) are inverse functions
<u>case D)</u>
Find the inverse of f(x)
Let
y=f(x)
Exchange variables x for y and y for x
Isolate the variable y
Let
therefore
f(x) and g(x) is not a pair of inverse functions
Answer:
1)
Minimum is a 4th Degree
2)
Positive; Even
3)
Step-by-step explanation:
Part 1)
The minimum degree of our function will be 4.
Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.
So, it can be anything 1, 3, 5, 7, etc.
However, we will choose the lowest one, 1.
Next, we know that the graph <em>bounces off</em> at 3.
So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.
We choose the lowest one, 2.
Therefore, the minimum degree of our function will be 1+1+2 or 4.
Part 2)
The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.
Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.
This applies to all polynomials with even degrees.
Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.
Part 3)
This is similar to Part 1.
We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is:
We know that it<em> passes through</em> . So, these two factors must have an odd multiplicity.
However, since the graph <em>bounces off</em> , this factor must have an even multiplicity.
And we're done!