Graph that equation. If the graph pass the horizontal line test, it is an inverse equation (because the graph of an inverse function is just a symmetry graph with respect to the line y= x of a graph of a one-to-one function). If it is given f(x) and g(x) as the inverse of f(x), check if g(f(x)) = x and f(g(x)) = x. If you show that g(f(x)) = x and f(g(x)) = x, then g(x) is the inverse of f(x).
I will use the quotient rule here. d/dx(f(x)/g(x) = g(x)*f'(x) - f(x)*g'(x)/[g(x)]2 x3*1/x - ln(x)*3x2/(x3)2 x3/x - 3ln(x)x2/x6 x2 - 3ln(x)x2/x6 = - 3ln(x)/x4 =========
y=x+1 there for answer is 2
y equals x-4 plus 2 is the same as y = x-2. You just translate the graph of y=x, 2 units to the right, OR 2 down.
It is translated left 8 and up 4.
8 x 6 = 48
The graph of g(x) is the graph of f(x) shifted 6 units in the direction of positive x.
Why
Graph that equation. If the graph pass the horizontal line test, it is an inverse equation (because the graph of an inverse function is just a symmetry graph with respect to the line y= x of a graph of a one-to-one function). If it is given f(x) and g(x) as the inverse of f(x), check if g(f(x)) = x and f(g(x)) = x. If you show that g(f(x)) = x and f(g(x)) = x, then g(x) is the inverse of f(x).
x = 6 looks like a vertical line which is centred on the point (6,0)
I will use the quotient rule here. d/dx(f(x)/g(x) = g(x)*f'(x) - f(x)*g'(x)/[g(x)]2 x3*1/x - ln(x)*3x2/(x3)2 x3/x - 3ln(x)x2/x6 x2 - 3ln(x)x2/x6 = - 3ln(x)/x4 =========
If x equals a constant number, the graph will be a vertical line. For example, the graph of x = 5 would be a vertical line that goes through the point (5,0). x equals 5 on every point along this lines.
3
y=x+1 there for answer is 2
y equals x-4 plus 2 is the same as y = x-2. You just translate the graph of y=x, 2 units to the right, OR 2 down.
The graph shifts downward (negative y) by 9 units.
First, reflect the graph of y = x² in the x-axis (line y = 0) to obtain the graph of y = -x²; then second, shift it 3 units up to obtain the graph of y = -x² + 3.