Which interval for the graphed function contains the local maximum?
2 answers:
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The solution is the point of intersection between the two equations.
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
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.
