g(x) = x/2
12
f(x)=2x+2. Put in 0, 1/2, 1, 3/2, 2... and you will get integer values. That is for the domain. The numbers you get when you put that in are the range integral values.
3
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
What_is_the_area_bounded_by_the_graphs_of_fx_and_gx_where_fx_equals_xcubed_and_gx_equals_2x-xsquared
g(x) = x/2
True
please elaborate your question
y = 3x + 2x^2 + k where k is any number
f(x) = √(2x -3) f(x) = (2x - 3)^(1/2) f'(x) = (1/2)[(2x - 3)^(1/2 - 1)](2) f'(x) = (2x - 3)^(-1/2) f'(x) = 1/[(2x - 3)^(1/2)] f'(x) = 1/√(2x -3)
Domain is the number of x values that can be used and not cause an imaginary result. Range is the number of the y values that result. In f(x)=2x-5 the range is all real numbers.
12
5x lolzz 8)
f(x)=2x+2. Put in 0, 1/2, 1, 3/2, 2... and you will get integer values. That is for the domain. The numbers you get when you put that in are the range integral values.
The properties and conditions of FX to satisfy such that the fx equals its fourier seasons, should be of the fans, and the ones that actually watch the awseome series that are on FX. Lots of people are huge fans of FX and really get destroyed when there is no satisfaction of the fourier seasons.
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2