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Amari Flowers

Answered question

2021-10-29

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

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

Definition of continuity
Function is said to be continuous if

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)

Let a point

c \in [- 4, 4]

then,

lim_{x \to c^{-}} f (x) = lim_{x \to c^{-}} x \sqrt{16 - x^{2}}

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

And

lim_{x \to c^{+}} f (x) = lim_{x \to c^{+}} x \sqrt{16 - x^{2}}

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

And

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

So,

lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x) = f (c)

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

Jeffrey Jordon

Expert2022-06-24Added 2605 answers

Answer is given below (on video)

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