How do you change this into simplest form?
1 answer:
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First, recall that
So if , then
.
Second,
We know that and
, which means we should also have
.
Third,
but as we've already shown, we need to have , so we pick the negative root.
Finally,
Unfortunately, none of the given answers match, so perhaps I've misunderstood one of the given conditions... In any case, this answer should tell you everything you need to know to find the right solution from the given options.
Answer:
See attached for graphs
g(x) -- domain: -∞ < x < ∞; range: 0 < y < ∞
g^-1(x) -- domain: 0 < x < ∞; range: -∞ < y < ∞
Step-by-step explanation:
g(x) is an exponential decay function. Its base is 1/3, so each increase of 1 unit in x will multiply the y-value by a factor of 1/3. The graph will rapidly approach its horizontal asymptote of y=0 as x gets large. The y-intercept is (0, 1). Just as y gets smaller as x increases, so it gets larger as x decreases. Each decrease of x by 1 unit causes the y-value to be multiplied by 3.
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The graph of g^-1(x) is the graph of g(x) reflected across the line y=x. That is, each coordinate pair (x, y) on the graph of g(x) becomes a point (y, x) on the graph of the inverse function. In order to graph g^-1(x), you don't need to write down the function, you only need to know the relationship between the graphs.
Just as x- and y- are interchanged on the graph, so the domain, range, and intercepts are interchanged. g^-1(x) will have a vertical asymptote of x=0, and an x-intercept of (1, 0). The domain of g^-1(x) is the range of g(x): 0 < x < ∞; and the range of g^-1(x) is the domain of g(x): -∞ < y < ∞.
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The attached graph shows g(x) in red and g^-1(x) in blue. As you can see, we created the graph simply by interchanging x and y. The line y=x is shown for reference, so you can see that each curve is a reflection of the other across that line.
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<em>Additional comment</em>
The explicit expression for g^-1(x) can be found by solving for y:
x = g(y)
If you're familiar with the log function, you know it has an x-intercept of 1 and a vertical asymptote at x=0. The base of the log function is simply a vertical scale factor. The minus sign reflects it across the x-axis.
