Let
f
(
x
)
=
3
x
5
and
g
(
x
)
=
3
x
2
4
x
f(x) = x2 + 3
g(x) = x + 3
f(g(x)) = g(x)2 + 3
= (x + 3)2 + 3
= x2 + 6x + 12
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
3.
f(x) = x2 + 3 ----> f(5) = (5)2 + 3 ----> f(5) = 28
f(x) = 3x2 + 5x + 2fprime(x) = 6x + 5
y - 8 = x^2 + 12x + 36 y - 8 = (x + 6)^2 y = (x + 6)^2 + 8 or f(x) = (x + 6)^2 + 8 So -h = 6 so that h = -6, and k = 8
14x
F(x)=[x^2]+1
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
F(x) = 15x2 - 2.5 + 3 That's a quadratic or 2nd degree polynomial in x.
f(x) = x2 + 5x + 1 The roots of this equation are x = -0.2087 and x = -4.7913 (approx).
If: df+10f = 3 Then: f(d+10) = 3 And: d = 3/f -10
No, by itself it is not.
f^2 + 2f = f (f + 2)
-2, 1.74 and 0.46
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
If: 5f+3 = 28 then f = 5 because (5*5)+3 = 28
wth??