If f(x) = 1/(x-1)
then
f-1(x) = (1/x) +1
PS...Type your question more clearly.
If it is the reciprocal of x minus 5 than 1/x-5>1/3x
The are under the curve on the domain (a,b) is equal to the integral of the function at b minus the integral of the function at a
-1/5
9
33/4
If it is the reciprocal of x minus 5 than 1/x-5>1/3x
The are under the curve on the domain (a,b) is equal to the integral of the function at b minus the integral of the function at a
Do you even know what a function is, stupid. A function is an equation that passes the vertical line test. go graph it bitc h
-1/5
To shift a funcion (or its graph) down "a" units, you subtract "a" from the function. For example, x squared gives you a certain graph; "x squared minus a" will give you the same graph, but shifted down "a" units. Similarly, you can shift a graph upwards "a" units, by adding "a" to the function.
No. That function describes a parabola who's vertex is at the point (0, -4).
9
The reciprocal of a number is 1 divide by that number - the reciprocal of -3 is 1/(-3), or -1/3. So, the negative reciprocal of -3 would be 1/3 (without the minus sign).
The graph of y = log(x) is defined only for x>0. The graph is a monotonic increasing function over its domain. It starts from an asymptotic "minus infinity" when x approaches 0. It passes through the value y = 0 when x = 1. The graph is illustrated at the link below.
a minus and a plus equal a minus number yeh dude
A bigger minus.
Assuming the standard x and y axes, the range is the maximum value of y minus minimum value of y; and the domain is the maximum value of x minus minimum value of x.