Lets explore function composition and the more sinister, decomposition!
Function composition is when one function is evaluated by the output of another function. For example, \( f( x+h ) \) can be viewed as a composite function where the outside function is \( f(x) \) and the inside function is \( g(x)=x+h \). Notice \( f(x+h) = f(g(x)) \).
Thats fun and all, but the trick for us will be to go backwards (which I call "decomposition" because I think the play on words is funny). Can you think of two functions that might be hiding in the expression, \( (x^3-1)^2 \)?
For each function below, find \( a(b) \) and \( b(x) \) so that \( f(x) = a( b(x)) \). Also, this is calculus class so we might as well compute each derivative!
1. \( f(x) = (x^2+x+1)^3 \)
\( a(b) \) \(= b^3 \) | \( a'(b) \) \(= 3 b^2 \) |
\( b(x) \) \(= x^2 + x + 1 \) | \( b'(x) \) \(= 2x + 1 \) |
2. \( f(x) = \sqrt{ 3x^3 -1 } \)
\( a(b) \) \(= \sqrt{b} = b^{\frac{1}{2}} \) | \( a'(b) \) \(= \frac{1}{2 \sqrt{b}} \) |
\( b(x) \) \(= 3x^3 - 1 \) | \( b'(x) \) \(= 9x^2 \) |
3. \( f(x) = e^{x^3} \)
\( a(b) \) \(= e^b \) | \( a'(b) \) \(= e^b \) |
\( b(x) \) \(= x^3 \) | \( b'(x) \) \(= 3 x^2 \) |
4. \( f(x) = \sin(2x^3+x) \)
\( a(b) \) \(= \sin(b) \) | \( a'(b) \) \(= \cos(b) \) |
\( b(x) \) \(= 2x^3 + x \) | \( b'(x) \) \(= 6x + 1 \) |
5. Compute each \( f'(x) \) above using WolframAlpha. How is each \(f'(x)\) is related to \( a(b), b(x), a'(b), b'(x) \)?
Each should be equivalent to \( a'( b ) \cdot b'(x) \), which can also be written as \( a'( b(x) ) \cdot b'(x) \).