D(x)=x^2+2x
C={(x,y)|y=4x-5}
1) D(C(x))
f(x)=1+3 sqrt of x+2
2) f(f(x))
D(x) = x^2+2x
D(C) = C^2+2C
= (4x-5)^2 + 2(4x-5)
= 16x^2 - 32x + 15
f(x) = 1+√(x+2)
f(f) = 1+√(f+2)
= 1+√(1+√(x+2)+2)
= 1+√(3+√(x+2))
D(C(x))
= D(4x-5)
= (4x-5)^2 + 2(4x-5)
expand and simplify if necessary
f(x) = 1 + 3√(x+2)
f(f(x))
= 1 + 3√(x+2)
= 1 + 3√((1 + 3√(x+2))+2)
= 1 + 3√(3 + 3√(x+2) )
testing:
let x = 2
f(2) = 1 + 3√4 = 7
f(7) = 1 + 3√9 = 10
f(f(2)) in my answer
= 1 + 3√(3+3√4)
= 1 + 3√9
= 1+9
= 10
there is a high probability that my answer is correct
I misread the 2nd problem - go with Reiny.
To find D(C(x)), we need to substitute the equation of C(x) into the equation of D(x).
1) D(x) = x^2 + 2x
C(x) = {(x, y) | y = 4x - 5}
Substituting y = 4x - 5 into D(x), we get:
D(C(x)) = (4x - 5)^2 + 2(4x - 5)
Simplifying further:
D(C(x)) = 16x^2 - 40x + 25 + 8x - 10
Combining like terms:
D(C(x)) = 16x^2 - 32x + 15
Therefore, D(C(x)) = 16x^2 - 32x + 15.
For the second question, we are given the function f(x) = 1 + 3√x + 2. We need to find f(f(x)).
2) f(x) = 1 + 3√x + 2
To find f(f(x)), we substitute f(x) into f(x).
f(f(x)) = 1 + 3√(1 + 3√x + 2) + 2
Simplifying further, we substitute x into the equation:
f(f(x)) = 1 + 3√(1 + 3√(x) + 2) + 2
Now, we have found f(f(x)).