if f(x) = 3x - 10, then whatever is put (substituted) for x in the "f(x)" bit is substituted for x in the "3x - 10" bit.
Thus f(2a) = 3(2a) - 10 = 6a - 10.
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
At x = 3, the value of F(x) = 3x + 2 is the value 11, which graphs to the point (3, 11).
What answer is this f(x)=3x+10 and g(x)=2x-4 find (f+g)(x)?
y = 3x - 4 or f(x) = 3x - 4 f(50) = 3(50) - 4 = 150 - 4 = 146 Thus, the y coordinate of the point is 146.
3x+5
fx=3x-5 f=(3x-5)/x f2=(6x-10)/x /=divide
y = f(x) = 2x2 - 3x + 10dy = (4x - 3) dxNote:It's possible that the function in the question may have been f(x) = (2x)2 - 3x + 10 .If that's the case, thendy = (8x - 3) dx
Let's simplify: f(x) = 3x-3x, so f(x) =0, so y=0 The graph is a line along the x-axis.
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
If f(x) = 2 - 3x then f(4) = 2 - 3(4) so since 3 times 4 is 12, then f(4) = 2 - 12 which means f(4) = -10 Easy!
f(x) = (3x - 4)/5 f(0) = [3(0) - 4]/5 = -4/5
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
y = 2x2 + 3x + 6 Since a > 0 (a = 2, b = 3, c = 6) the graph opens upward. The coordinates of the vertex are (-b/2a, f(-b/2a)) = (- 0.75, 4.875). The equation of the axis of symmetry is x = -0.75.
the linearization of f(x) = x^4 + 3x^2 with a = -1 is L(x) = 4 - 10(x+1) the linearization formula is L(x) = f(a) + f'(a)*(x-a) when, a = -1 f(a) = (-1)^4 + 3(-1)^2 = 4 f'(x) = 4*x^3 + 6*x f'(a) = 4(-1)^3 + 6(-1) = -10 so, L(x) = (4) + (-10)*(x-(-1)) = 4 - 10(x+1)
At x = 3, the value of F(x) = 3x + 2 is the value 11, which graphs to the point (3, 11).
if f(x) = 3x + 2 and you're looking for f(2), just plug in 2 for x: f(2) = 3(2) + 2 = 6 + 2 = 8.
What answer is this f(x)=3x+10 and g(x)=2x-4 find (f+g)(x)?