Amari Flowers

2021-10-29

Use the definition of continuity and the properties of limits to show that the function

$f\left(x\right)=x\sqrt{16-{x}^{2}}$ is continuous on the interval [-4,4]

Nathaniel Kramer

Skilled2021-10-30Added 78 answers

Consider the given function as

$f\left(x\right)=x\sqrt{16-{x}^{2}}$

Definition of continuity

Function is said to be continuous if$\underset{x\to {a}^{-}}{lim}f\left(x\right)=\underset{x\to {a}^{+}}{lim}f\left(x\right)=f\left(a\right)$

Let a point$c\in [-4,4]$ then,

$\underset{x\to {c}^{-}}{lim}f\left(x\right)=\underset{x\to {c}^{-}}{lim}x\sqrt{16-{x}^{2}}$

$=c\sqrt{16-{c}^{2}}$

And

$\underset{x\to {c}^{+}}{lim}f\left(x\right)=\underset{x\to {c}^{+}}{lim}x\sqrt{16-{x}^{2}}$

$=c\sqrt{16-{c}^{2}}$

And

$f\left(c\right)=c\sqrt{16-{c}^{2}}$

So,$\underset{x\to {c}^{-}}{lim}f\left(x\right)=\underset{x\to {c}^{+}}{lim}f\left(x\right)=f\left(c\right)$

Hence, function is continuous on the given interval [-4,4]

Definition of continuity

Function is said to be continuous if

Let a point

And

And

So,

Hence, function is continuous on the given interval [-4,4]

Jeffrey Jordon

Expert2022-06-24Added 2607 answers

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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