True
y = 3x + 2x^2 + k where k is any number
f(x) = (2x^2 + 5x - 12)/(x + 4) as given, is not defined when the denominator is 0 ie when x + 4 = 0 or x = -4. However, 2x^2 + 5x - 12 = (x + 4)*(2x - 3) so that the function can be simplified to (x + 4)*(2x - 3)/(x + 4) = 2x - 3 This has a zero when 2x - 3 = 0 ie 2x = 3 or x = 1.5
You need to clarify the function AND provide an interval.
[fx] is a function of x, it usually used in graphs.
12
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
5x lolzz 8)
Def FX ended in 1997.
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)
True
f(x) is the same thing as y= example: f(x)=2x+3 OR y=2x+3
please elaborate your question
y = 3x + 2x^2 + k where k is any number
4
first of all, what is a plu? but the answer is (-8)+f+1/x sqared by the way, what grade are you in?