Diketahui f (x) = x² + 6x + 5 , g(x) = √x+4 , dan h(x) = x² - 5x + 1 . rumus fungsi ( g ° f ° h )(x) = ...
Takamori37
Tentukan:
![$\begin{align}(g\circ f\circ h)(x)&=g(f(h(x))) \\ &=g([h(x)]^2+6[h(x)]+5) \\ &=\sqrt{[h^2(x)+6h(x)+5]+4} \\ &=\sqrt{h^2(x)+6h(x)+9} \\ &=\sqrt{(h(x)+3)^2} \\ &=h(x)+3 \\ &=(x^2-5x+1)+3 \\ &=x^2-5x+4\end{align}](https://tex.z-dn.net/?f=%24%5Cbegin%7Balign%7D%28g%5Ccirc+f%5Ccirc+h%29%28x%29%26%3Dg%28f%28h%28x%29%29%29+%5C%5C+%26%3Dg%28%5Bh%28x%29%5D%5E2%2B6%5Bh%28x%29%5D%2B5%29+%5C%5C+%26%3D%5Csqrt%7B%5Bh%5E2%28x%29%2B6h%28x%29%2B5%5D%2B4%7D+%5C%5C+%26%3D%5Csqrt%7Bh%5E2%28x%29%2B6h%28x%29%2B9%7D+%5C%5C+%26%3D%5Csqrt%7B%28h%28x%29%2B3%29%5E2%7D+%5C%5C+%26%3Dh%28x%29%2B3+%5C%5C+%26%3D%28x%5E2-5x%2B1%29%2B3+%5C%5C+%26%3Dx%5E2-5x%2B4%5Cend%7Balign%7D "$\begin{align}(g\circ f\circ h)(x)&=g(f(h(x))) \\ &=g([h(x)]^2+6[h(x)]+5) \\ &=\sqrt{[h^2(x)+6h(x)+5]+4} \\ &=\sqrt{h^2(x)+6h(x)+9} \\ &=\sqrt{(h(x)+3)^2} \\ &=h(x)+3 \\ &=(x^2-5x+1)+3 \\ &=x^2-5x+4\end{align}")
4 votes
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