work on better figures
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jverzani
@@ -30,20 +30,57 @@ Why is this useful? When available, it can help us solve equations. If we can wr
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Let's explore when we can "solve" for an inverse function.
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Consider the graph of the function $f(x) = 2^x$:
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Consider this graph of the function $f(x) = 2^x$
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```{julia}
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#| hold: true
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f(x) = 2^x
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plot(f, 0, 4, legend=false)
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plot!([2,2,0], [0,f(2),f(2)])
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#| echo: false
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p = let
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gr()
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empty_style = (xaxis=([], false),
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yaxis=([], false),
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framestyle=:origin,
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legend=false)
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axis_style = (arrow=true, side=:head, line=(:gray, 1))
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text_style = (10,)
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fn_style = (;line=(:black, 3))
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fn2_style = (;line=(:red, 4))
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mark_style = (;line=(:gray, 1, :dot))
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domain_style = (;fill=(:orange, 0.35), line=nothing)
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range_style = (; fill=(:blue, 0.35), line=nothing)
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f(x) = 2^x
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a, b = 0, 2
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plot(; empty_style...)
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xs = range(a, b, 200)
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ys = f.(xs)
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plot!( xs, ys; fn_style...)
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plot!([a-1/4, b+.2], [0,0]; axis_style...)
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plot!([0, 0], [-.1, f(2.1)]; axis_style...)
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x = 1
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y = (f(b)+f(a))/2
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plot!([x,x,0],[0,f(x),f(x)]; line=(:black, 1, :dash), arrow=true, side=:head)
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plot!([0,log2(y), log2(y)], [y,y,0]; line=(:black,1,:dash), arrow=true, side=:head)
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annotate!([
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(x, 0, text(L"c", 10, :top)),
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(0,f(x), text(L"f(c)", 10, :right)),
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(0, y, text(L"y=f(d)", 10, :right)),
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(log2(y), 0, text(L"d", 10, :top))
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])
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end
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plotly()
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p
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```
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The graph of a function is a representation of points $(x,f(x))$, so to *find* $y = f(c)$ from the graph, we begin on the $x$ axis at $c$, move vertically to the graph (the point $(c, f(c))$), and then move horizontally to the $y$ axis, intersecting it at $y = f(c)$. The figure shows this for $c=2$, from which we can read that $f(c)$ is about $4$. This is how an $x$ is associated to a single $y$.
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If we were to *reverse* the direction, starting at $y = f(c)$ on the $y$ axis and then moving horizontally to the graph, and then vertically to the $x$-axis we end up at a value $c$ with the correct $f(c)$. This allows solving for $x$ knowing $y$ in $y=f(x)$.
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If we were to *reverse* the direction, starting at $y = f(d)$ on the $y$ axis and then moving horizontally to the graph, and then vertically to the $x$-axis we end up at a value $d$ with the correct $f(d)$. This allows solving for $x$ knowing $y$ in $y=f(x)$.
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The operation described will form a function **if** the initial movement horizontally is guaranteed to find *no more than one* value on the graph. That is, to have an inverse function, there can not be two $x$ values corresponding to a given $y$ value. This observation is often visualized through the "horizontal line test" - the graph of a function with an inverse function can only intersect a horizontal line at most in one place.
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@@ -309,6 +346,131 @@ What do we see? In blue, we can see the familiar square root graph along with a
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This is reminiscent of the formula for the slope of a perpendicular line, $-1/m$, but quite different, as this formula implies the two lines have either both positive slopes or both negative slopes, unlike the relationship in slopes between a line and a perpendicular line.
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::: {#fig-inverse-normal layout-ncol=1}
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```{julia}
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#| echo: false
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# inverse function slope
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gr()
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p1 = let
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f(x) = x^2
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df(x) = 2x
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empty_style = (xaxis=([], false),
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yaxis=([], false),
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framestyle=:origin,
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legend=false)
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axis_style = (arrow=true, side=:head, line=(:gray, 1))
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text_style = (10,)
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fn_style = (;line=(:black, 3))
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fn2_style = (;line=(:red, 4))
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mark_style = (;line=(:gray, 1, :dot))
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plot(; aspect_ratio=:equal, empty_style...)
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xs = range(0, 1.25, 100)
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plot!(xs,f.(xs); fn_style...)
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plot!(f.(xs), xs; fn2_style...)
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plot!(identity, -1/4, 2; line=(:gray, 1, :dot))
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#plot!([-.1, 1.35],[0,0]; axis_style...)
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#plot!([0,0], [-0.1, f(1.3)]; axis_style...)
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c = .4
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m = df(c)
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tl(x) = f(c) + df(c)*(x-c)
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plot!(tl; line=(:black, 1, :dash))
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d = c + .6
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p1, p2, p3 = (c, tl(c)), (d, tl(c)), (d, tl(d))
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q1, q2, q3 = (tl(c),c), (tl(c),d), (tl(d), d)
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plot!([p1, p2, p3]; line=(:black, 1, :dot))
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tl1(x) = c + (1/m)*(x - f(c))
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plot!(tl1; line=(:red, 1, :dash))
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plot!([q1, q2, q3]; line=(:red, 1, :dot))
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annotate!([
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((c+d)/2, f(c), text(L"\Delta x", 10, :top, :black)),
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(d, (tl(c)+tl(d))/2, text(L"\Delta y", 10, :left, :black)),
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(f(c), (c+d)/2, text(L"\Delta x", 10, :right, :red)),
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((tl(c)+tl(d))/2, d, text(L"\Delta y", 10, :bottom, :red)),
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(d, tl(d),
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text(L"rise/run = $m = \Delta y / \Delta x$", 10, :top, :left,
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rotation= rad2deg(atan(m)))),
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(tl(d), d,
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text(L"rise/run = $\Delta x / \Delta y = 1/m$", 10, :bottom, :left, rotation=rad2deg(atan(1/m)))),
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(1.9, 1.9, text(L"y=x", 10, :top, rotation=45))
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])
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current()
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end
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# normal line
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p2 = let
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f(x) = 4 - (x-2)^2
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df(x) = -2*(x-2)
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empty_style = (xaxis=([], false),
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yaxis=([], false),
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framestyle=:origin,
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legend=false)
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axis_style = (arrow=true, side=:head, line=(:gray, 1))
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text_style = (10,)
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fn_style = (;line=(:black, 3))
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fn2_style = (;line=(:red, 4))
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mark_style = (;line=(:gray, 1, :dot))
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plot(; aspect_ratio=:equal, empty_style...,
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xlims=(1, 2.5), ylims=(3, 4.5))
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xs = range(.99, 2.01, 100)
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plot!(xs,f.(xs); fn_style...)
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c = 1.5
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tl(x) = f(c) + df(c)*(x-c)
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nl(x) = f(c) - (x-c)/df(c)
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xs = range(1, 2, 10)
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plot!(xs, tl.(xs); fn2_style...)
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xs = range(.9, 2, 10)
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plot!(xs, nl.(xs); fn2_style...)
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ylims!((f(.85), nl(.95)))
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o = 1/3
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plot!([c,c+o, c+o], [tl(c), tl(c), tl(c+o)]; mark_style...)
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m = (tl(c+o) - tl(c))
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plot!([c,c,c+m], [nl(c),nl(c + m),nl(c+m)]; mark_style...)
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theta = rad2deg(atan(tl(c+o)-tl(c), o))
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annotate!([
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(c + o/2, f(c), text(L"1", :top, 10)),
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(c + o, (f(c)+f(c+o))/2, text(L"m", :right, 10)),
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(c, (nl(c) + nl(c+m))/2, text(L"-1", :right, 10)),
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(c+m/2, nl(c+m), text(L"m", :top, 10)),
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(c + o/2, tl(c+o), text(L"rise/run $=m/1$", 10, :top,
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rotation=theta)),
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(c + 1.1*o, nl(c+1.1*o), text(L"rise/run $=(-1)/m$", 10, :bottom,
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rotation=theta-90))
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])
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current()
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end
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plot(p1, p2)
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```
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The inverse function has slope at a corresponding point that is the *reciprocal*; the "normal line" for a function at a point has slope that is the *negative reciprocal* of the "tangent line" at a point.
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:::
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```{julia}
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#| echo: false
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plotly()
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nothing
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```
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The key here is that the shape of $f(x)$ near $x=c$ is somewhat related to the shape of $f^{-1}(x)$ at $f(c)$. In this case, if we use the tangent line as a fill in for how steep a function is, we see from the relationship that if $f(x)$ is "steep" at $x=c$, then $f^{-1}(x)$ will be "shallow" at $x=f(c)$.
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