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
10 to the power of minus 1 is equal to 1 divided by 10, which is 0.1. This can be calculated by taking the reciprocal of 10 raised to the power of 1, which is 10. Therefore, 10 to the power of minus 1 is equivalent to 0.1.
9
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
10 to the power of minus 1 is equal to 1 divided by 10, which is 0.1. This can be calculated by taking the reciprocal of 10 raised to the power of 1, which is 10. Therefore, 10 to the power of minus 1 is equivalent to 0.1.
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).
A bigger minus.
a minus and a plus equal a minus number yeh dude
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.