I know that a shift left by a means f(x+a). Then a y-axis reflection gives f(-(x+a)). Finally, shifting right again gives f(-x-a-a) = f(-2a-x). However, the answers stipulate f(2a-x), as shown. What am I doing wrong?
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When reflecting across the $y$-axis, only the variables $x$ must be negated, not the whole argument to $f$. So it should be $f(-x+a)$.
Parcly Taxel
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I've edited the question to provide more context. What would you mean by shift the axes right? – user71207 Oct 27 '20 at 06:34
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ahhh I see. That may explain it. A very weird approach though. @A.J pointed out my mistake though anyway – user71207 Oct 27 '20 at 06:48
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I don't understand how the book is defining its transformations so I'm ignoring it. As far as your steps go, you're just making a mistake in the second step.
Step 1 - shift left by $a$: to do this, we replace $x$ with $x+a$. So
$$f(x) \longrightarrow f(x+a)$$
Step 2 - reflect in $y$-axis: to do this, we replace $x$ with $-x$. So
$$f(x+a) \longrightarrow f(-x+a)$$
Step 3 - shift right by $a$: to do this, we replace $x$ with $x-a$. So
$$f(-x+a) \longrightarrow f\left[-(x-a)+a\right]=f(2a-x)$$
A.J.
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Thank you. Could you explain part b)iii) as well? I get$ g(x)$ $-->$ $g(2a -x)$, but I;m stuck on proving the result using x = a + t. If I sub that in, g(2a - a - t) = g(a - t). Also, from g(x) $-->$ g(a+t). This is not the same? – user71207 Oct 27 '20 at 06:56
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If $g(x)=g(2a-x)$, then $g(x)$ is symmetric about the line $x=a$. What you did is exactly correct, you just need to consider what it means that $g(a-t)=g(a+t)$ for every value of $t$. – A.J. Oct 27 '20 at 07:17
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Hmm I don't understand what it means, that's the problem. One shifts left and one shifts right by "t" units. If they are equal... I can't find a way to explain it. Any hints? Symmetry pops into my mind but I'm imagining a line like y = x being moved across the x axis, and I dont know how any 2 shifts in opposite directions could be the same – user71207 Oct 27 '20 at 07:24
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1You need to look at it not as a transformation. Instead, think of it like this: $a-t$ is the $x$-coordinate $t$ units to the left of $a$, while $a+t$ is the $x$-coordinate $t$ units to the right of $a$. So the fact that $g(a-t)=g(a+t)$ means that those two points share the same $y$-coordinate; so those two points at least are reflections of each other across the line $x=a$. Now if this is true for every value of $t$, then $g(x) $ must be symmetric about the line $x=a$. – A.J. Oct 27 '20 at 22:57