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Place the cursor over the image to start the animation.
The cardioid from Problem 6 |
The cardioid from Problem 6 scaled by 2 |
The cardioid from Problem 6 scaled by 2 and translated. |
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RegionPlot3D[
And[y^2 + z^2 < 1, z^2 + x^2 < 1], {x, -1, 1}, {y, -1, 1}, {z, -1,
1}, PlotPoints -> {200, 200, 200}, Mesh -> None,
PlotStyle -> {Opacity[1]},
Ticks -> {Range[-2, 2, 1], Range[-2, 2, 1], Range[-2, 2, 1]},
ImageSize -> 500
]
RegionPlot3D[
And[x^2+y^2 < 1, y^2 + z^2 < 1, z^2 + x^2 < 1], {x, -1, 1}, {y, -1, 1}, {z, -1,
1}, PlotPoints -> {200, 200, 200}, Mesh -> None,
PlotStyle -> {Opacity[1]},
Ticks -> {Range[-2, 2, 1], Range[-2, 2, 1], Range[-2, 2, 1]},
ImageSize -> 500
]
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A cross-section is an equilateral triangle.
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A cross-section is a square.
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A cross-section is a regular pentagon.
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A cross-section is a regular hexagon.
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A cross-section is a regular heptagon.
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A cross-section is a regular octagon.
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Folium of Decartes $x^3+y^3 = 6xy$ |
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The functions $x^2$ and $\sqrt{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
The functions $x^3$ and $\sqrt[3]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
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The functions $x^4$ and $\sqrt[4]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
The functions $x^5$ and $\sqrt[5]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
Figure for $\displaystyle \lim_{x\to 0} \frac{1- \cos x}{x^2} = \frac{1}{2}$ |
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The functions $x^2$ and $\sqrt{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
The functions $x^3$ and $\sqrt[3]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
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The functions $x^4$ and $\sqrt[4]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
The functions $x^5$ and $\sqrt[5]{x}$ on $[0,1]$ and the maximal disk, with the oversized center, that touches both graphs. |
the navy blue function is $e^x$, the purple circle is a unit circle
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the sine function in navy blue and its many tangents in gray
$ X =\mathbb R, \ Y = (0,+\infty), \ f = \bigl\{ \bigl(x, xe^x/(e^x-1) \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $ The function $f$ from the example above |
$X = \mathbb R, \ Y = (0,+\infty), \ g = \bigl\{ \bigl(x,(e^x - 1)/x \bigr) \ \bigl| \bigr. \ x \in X \bigr\}$ The function $g$ from the example above |
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$ X = [-1,+\infty), \ Y = [-1/e,+\infty), \ W_0 = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $ The Lambert $W_0$ function |
$X = (-\infty,-1), \ Y = (-1/e,0), \ W_{-1} = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\}$ The Lambert $W_{-1}$ function |
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for every real number $y$ between $f(a)$ and $f(b)$ there exists $x \in [a,b]$ such that $y = f(x)$.
there exist $x_0, x_1 \in [a,b]$ such that $f(x_0) \leq f(x) \leq f(x_1)$ for all $x \in [a,b]$.
$|x-a| \lt \delta(\epsilon)$ implies $|f(x) - f(a) | \lt \epsilon$.
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x-2| \lt \delta(\epsilon)$ implies $|x^2 - 4 | \lt \epsilon$.
$0 \lt \delta(\epsilon) \leq 1$ and $0 \lt |x-2| \lt \delta(\epsilon)$ implies $|x^2 - 4 | \lt \epsilon$.
$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x-0| = |x| \lt \delta(\epsilon)$ implies $|\cos x -1 | \lt \epsilon$.
$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $|\cos x - 1 | \lt \epsilon$.
Figure for $\displaystyle \lim_{x\to 0} \cos x = 1$ |
Figure for $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$ |
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$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x-0| = |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon$.
$0 \lt \delta(\epsilon) \leq \pi/2$ and $0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon$.
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The composition at a generic point $x$ |
The composition at another generic point $x$ |
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Finding a minimum of the composition |
Finding a maximum of the composition |
$0 \lt |x-a| \lt \delta(\epsilon)$ implies $|f(x) - L | \lt \epsilon$.
The sign function |
The unit step function |
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The floor function |
The ceiling function |
$ X = [0,+\infty), \ Y = [0,+\infty), \ f_1 = \bigl\{ (x,x^2) \ \bigl| \bigr. \ x \in X \bigr\} $ |
$X = \mathbb R, \ Y = \mathbb R, \ f_2 = \bigl\{ (x,x^3) \ \bigl| \bigr. \ x \in X \bigr\}$ |
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$X= \mathbb R, \ Y = (0,+\infty), \ f_3 = \bigl\{ (x,e^x) \ \bigl| \bigr. \ x \in X \bigr\}$ |
$X = (-\pi/2,\pi/2), \ Y = \mathbb R, \ f_4 = \bigl\{ \bigl( x,\tan(x) \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $ |
$ X = [-\pi/2,\pi/2], \ Y = [-1,1], \ f_5 = \bigl\{ \bigl(x,\sin(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} $ |
$ X = [0,\pi], \ Y = [-1,1], \ f_6 = \bigl\{ \bigl(x,\cos(x)\bigr) \ \bigl| \bigr. \ x \in X \bigr\} $ |