If f(x)=x2+10, then f(x+h)=?
f(x+h)=(x+h)2+10 (since f(x)=x2+10, substitute the x in x2 to (x+h)2)=(x+h)(x+h)+10 (then multiply (x+h) by (x+h) by doing the FOIL method)=x2+xh+xh+h2+10 (First: x*x, Outside: x*h, Inside: h*x, Last: h*h)=x2+2xh+h2+10 (combine like terms (xh+xh=2xh))
So if f(x)=x2+10, then f(x+h)=x2+2xh+h2+10
73984637247673673u7489348873q9unvhn
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
s=5 f+s=? Substitute 5 in for s.Substitute 1 in for f.f+s=f+5f=11+5 ={6}
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.
They are at x = -3 and x = -2.
f(x)= (4 - 5y)/8 f(y)= (4 - 8x)/5
14x
F(x)=[x^2]+1
f+7=12 is the equation so, f=5
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
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
F = 9/5 C + 32 = (9/5) x 10 + 32 = 18 + 32 = 50
f^2 + 2f = f (f + 2)
-2, 1.74 and 0.46
wth??
yup
So what is the question?