daviddoji/CalculusWithJuliaNotes.jl
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some typos

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Fang Liu
2025-04-22 15:19:44 +08:00
parent 36895faafe
commit ed1d92197a
8 changed files with 20 additions and 20 deletions
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6
quarto/precalc/inversefunctions.qmd
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@@ -182,7 +182,7 @@ x &= \sqrt{(1-y)/y}, \quad 0 < y \leq 1.
\end{align*}
$$
Then $f^{-1}(x) = \sqrt{(1-x)/x}$ where $0 < x \leq 1$. The somewhat complicated restriction for the the domain coincides with the range of $f(x)$. We shall see next that this is no coincidence.
Then $f^{-1}(x) = \sqrt{(1-x)/x}$ where $0 < x \leq 1$. The somewhat complicated restriction for the domain coincides with the range of $f(x)$. We shall see next that this is no coincidence.
## Formal properties of the inverse function
@@ -198,7 +198,7 @@ plot(f, 0, 4, legend=false)
plot!([2,2,0], [0,f(2),f(2)])
```
The graph is shown over the interval $(0,4)$, but the *domain* of $f(x)$ is all $x \geq 0$. The *range* of $f(x)$ is clearly $2 \leq x \leq \infty$.
The graph is shown over the interval $(0,4)$, but the *domain* of $f(x)$ is all $x \geq 0$. The *range* of $f(x)$ is clearly $2 \leq y \leq \infty$.
The lines layered on the plot show how to associate an $x$ value to a $y$ value or vice versa (as $f(x)$ is one-to-one). The domain then of the inverse function is all the $y$ values for which a corresponding $x$ value exists: this is clearly all values bigger or equal to $2$. The *range* of the inverse function can be seen to be all the images for the values of $y$, which would be all $x \geq 0$. This gives the relationship:
@@ -283,7 +283,7 @@ The slope of $f(x) = 9/5 \cdot x + 32$ is clearly $9/5$ and the slope of the inv
Now consider the graph of the *tangent line* to a function. This concept will be better defined later, for now, it is a line "tangent" to the graph of $f(x)$ at a point $x=c$.
For concreteness, we consider $f(x) = \sqrt{x}$ at $c=2$. The tangent line will have slope $1/(2\sqrt{2})$ and will go through the point $(2, f(2)$. We graph the function, its tangent line, and their inverses:
For concreteness, we consider $f(x) = \sqrt{x}$ at $c=2$. The tangent line will have slope $1/(2\sqrt{2})$ and will go through the point $(2, f(2))$. We graph the function, its tangent line, and their inverses:
```{julia}