svrstanojpkqx

2022-03-25

Consider the function $f\left(x\right)=\sqrt{1-{x}^{2}}$ on the interval [0, -1], how do you find the average or mean slope of the function on this interval?

Demetrius Kaufman

The slope of a function f(x) at any point is equal to $\frac{d}{dx}\left(f\left(x\right)\right)$.
The average value of a function g(x) from $x\in \left[a,b\right]$ is given by $\frac{1}{b-a}{\int }_{a}^{b}g\left(x\right)dx$.
We want to find the average slope of the function $f\left(x\right)=\sqrt{1-{x}^{2}}$ on $x\in \left[-1,0\right]$.
Its slope at any given point is $\frac{d}{dx}\left(\sqrt{1-{x}^{2}}\right)$. Thus, the average slope from
x=-1 to x=0 is given by
$\frac{1}{0-\left(-1\right)}{\int }_{-1}^{0}\frac{d}{dx}\left(\sqrt{1-{x}^{2}}\right)dx$
By the Fundamental Theorem of Calculus, we know that integration and differentiation are inverse operations. Thus, we have the above expression equal to
$={\left[\sqrt{1-{x}^{2}}\right]}_{-1}^{0}$
=1

umgebautv6v2

Note that the "interval [0, -1] should mean the set of all x with $0\le x$ and i will assume that you mean [-1,0].
The slope is the same as the rate of change.
So the average slope on [-1, 0] is equal to the average rate of change on the interval:
$\frac{f\left(0\right)-f\left(-1\right)}{0-\left(-1\right)}=\frac{\sqrt{1}-\sqrt{0}}{1}=1$
If the intended interval is [0,1], then the average slope is
$\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{\sqrt{0}-\sqrt{1}}{1}=-1$

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