f(x)=x+1
g(f(x))=x
f(x)-1=x
g(x)=x-1
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
Let f ( x ) = 3 x 5 and g ( x ) = 3 x 2 4 x
d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)] or f'(x) + g'(x) when x = 3, d/dx [f(x) + g(x)] = f'(3) + g'(3) = 1.1 + 7 = 8.1 d/dx [f(x)*g(x)] = f(x)*d/dx[g(x)] + d/dx[f(x)]*g(x) when x = 3, d/dx [f(x)*g(x)] = f(3)*g'(3) + f'(3)*g(3) = 5*7 + 1.1*(-4) = 35 - 4.4 = 31.1
The browser used for posting questions on this site is rubbish. It is quite useless for mathematics since it strips out most symbols so that we cannot tell what the question is meant to be. I can only see f(x plus 1) 2x 1. I can see that it is means to be f(x+1) = 2x ~ 1but the ~ remains an unidentified operator.From the above, you can deduce that f(x) = 2(x - 1) ~1 = 2x - 2 ~1Similarly g(x) = 4x - 4 + 4 = 4x.Then f(g(x) = f(4x) = 2*4x - 2 ~ 1 = 8x - 2 ~ 1.You will need to substitute the appropriate operator for ~ and evaluate.
f x 3x plus 1 and g x x 2 -6 find fog 4 can not be solved.
g(x) = 5. So whatever f(x) or f(-1) is, g of that is going to be 5.
wth??
gf(5) = g(f(5)) = g(5+1) since f(x) = x+1 and then g(6) = 3*6 - 2 = 18 - 2 = 16
No, by itself it is not.
f(x) = x - 1g(x) = (x2 - 1)/(x + 1) = (x - 1 )(x + 1)/(x + 1) = x - 1therefore, f(x) = g(x)
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
1 plus 1 plus 1 plus 1 equals 1 times 4. 1 times 4 equals 4. 4 minus 4 equals 0. 0
Let f ( x ) = 3 x 5 and g ( x ) = 3 x 2 4 x
=b+2h+r+x+z+q+a+p+l+f+g+d
1 plus one
x = 75
z remains undefined.