f(x)= (x-7) /(-2x+7)(2x+5) set -2x+7=0 -2x=-7 x=7/2 set 2x+5=0 2x=-5 x=-5/2 x=7/2 and x=-5/2 are vertical asymptotes.
f(x) = 2x + 1 g(x) = x^2 - 7 So f*g(x) = f(g(x)) = f(x^2 - 7) = 2*(x^2 - 7) + 1 = 2*x^2 - 14 + 1 = 2*x^2 - 13
h(x)=3x-4
If 2x + 3y = 4, y= (4 - 2x)/3. In function notation, f(x) = (4 - 2x)/3.
You cannot because the function is not well-defined. There is no equality symbol, the function In(2x) is not defined.
f(x) = 2x + 8 f(x) = -7 → 2x + 8 = -7 → 2x = -15 → x = -7½ = -7.5
== == Suppose f(x) = x3 + 3x2 - 2x + 7 divisor is x + 1 = x - (-1); so rem = f(-1) = 11
f(x)= (x-7) /(-2x+7)(2x+5) set -2x+7=0 -2x=-7 x=7/2 set 2x+5=0 2x=-5 x=-5/2 x=7/2 and x=-5/2 are vertical asymptotes.
y=f(x)= x(2x+y)=7 f(x)=2x^2 +xy -7 = 0 y=2x^2 +xy -7 y-xy -7 + 2x^2 = 0 y(1-x)=2x^2-7 y=(2x^2-7) / (1-x) Excuse the working. Y equals 2 X squared minus 7 all divided by 1-X
f(x) = 2x + 1 g(x) = x^2 - 7 So f*g(x) = f(g(x)) = f(x^2 - 7) = 2*(x^2 - 7) + 1 = 2*x^2 - 14 + 1 = 2*x^2 - 13
The equation F X 2X plus 5 equals 0.5. This is a algebra math problem.
Subtract 6 from each side: 2x = 14; divide each side by 2: x = 7
f(x) = -2x + 1; g(x) = -4x; g(f(4)) = ? Solution: (g ° f)(4) = g(f(4)) = g(-7) = 28 f(x) = -2x + 1 f(4) = -2(4) + 1 f(4) = -7 g(x) = -4x g(-7) = -4(-7) g(-7) = 28
It is 5.
It is 5.
f(x) = 2x - 7g(x) = -4x + 6f.g(-2) = f(g(-2))g(-2) = -4*-2 + 6 = 8 + 6 = 14so f.g(-2) = f(14) = 2*14 - 7 = 28 - 7 = 21.
f+7=12 is the equation so, f=5