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mafiozo [28]
3 years ago
12

An exponential function f(x) is reflected across the x-axis to create the function g(x). Which is a true statement regarding f(x

) and g(x)? A. The the functions have the same initial value. B. The two functions will cross each other on the axis. C. The two functions have reciprocal output values of each other for any given output value. D. The two functions have opposite output values of each other for any given input value.

Mathematics
2 answers:
nlexa [21]3 years ago
9 0

Answer:

D

Step-by-step explanation: E2020

KIM [24]3 years ago
5 0

Answer:

<h2>D. The two functions have opposite output values of each other for any given input value.</h2>

Step-by-step explanation:

The reflect function will have opposite output values than the original one, because reflection across the x-axis means that y-values will be with the opposite sign. All positive y-values in the original function will have negative values in the reflected one, and all negative y-values in the original will have positive values in the reflected.

The opposite values are just seen in the output, because those belong to y-axis, all x-axis values remain the same, which are input values. You can see this in the image attached, look for the horizontal reflection. That mirror effect across the x-axis is because the y-values are opposite.

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4-7y=-2+6

4-7y=4

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y=0

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Your welcome! Have a great day!

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aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 2

\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 3

\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity \sin^2(\theta)+\cos^2(\theta) = 1 in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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