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$F: [0,π/2]\to \mathbb{R}$ where $F(x) = \text{sin(x)} + 77$ and $G: \mathbb{R} \to \mathbb{R}$ where $G(y) = y^4$ Show that $G(F(x))$ where domain is $[0,π/2]$ and codomain is $\mathbb{R}$ is injective.

I'm sorry for not writing this neatly. The way I went about this was to first show $F(x)$ is injective, then show $G(x)$ is not injective, then show $G(x)$ is injective over the domain $[0,π/2]$ since thats the domain of our composition function. Then proved that if $F(x)$ is 1-1 and $G(x)$ is 1-1 then $G(F(x))$ is 1-1. Is this correct?

Andrés Villa
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Hazard
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1 Answers1

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I mean, $g$ is injective on $\mathbb{R_+} := \{x \in \mathbb{R} \vert x\ge0\}$ and $f$ is injective on $[0,\frac{\pi}{2}]$ and $f([0,\frac{\pi}{2}]) \subset \mathbb{R_+}$, so that's enough to argue that $g(f(x))$ is injective, but if you want to show it with the definition:

Let $g(f(x_1)) = g(f(x_2))$. We want to show that this implies $x_1 = x_2$.

$g(f(x_1)) = g(f(x_2)) \iff (f(x_1))^4 = (f(x_2))^4 \iff f(x_1) = f(x_2)$ or $f(x_1) = -f(x_2)$.

However we have for all $x \in [0, \frac{\pi}{2}]$ that $f(x) = \sin(x) +77 \ge 77$ (since $\sin(x)$ is non-negative on $[0, \frac{\pi}{2}]$). So $77\le f(x_1) = -f(x_2)\le -77$ is impossible, meaning we have $$g(f(x_1)) = g(f(x_2)) \iff f(x_1) = f(x_2) \iff \sin(x_1) +77 = \sin(x_2) +77 \iff \sin(x_1) = \sin(x_2)$$ We know that $\sin$ is injective on $[0, \frac{\pi}{2}]$ (alternatively we know that $\sin$ is strictly increasing on this domain, which implies injectivity) so we finally get $\sin(x_1) = \sin(x_2) \iff x_1 = x_2$.

yagod
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