Question

Given $f\left(x\right)=\sqrt{9-x^2}$, then the function is continuous at [-3,3].

• A. True
• B. False

Answer 1

The correct option is A

True

A function f(x) is said to be continuous on a closed interval [a,b] if the condtions given below are

satisfied.

1) f(x) is continuous on (a,b)

2) f(x) is continuous from the right at 'a'

3) f(x) is continuous from the left at 'b'

In this case

$f\left(x\right)=\sqrt{9-x^2}$

for c in (-3,3)

$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}$ $\sqrt{9-x^2}=\sqrt{9-c^2}=f\left(c\right)$

So f(x) is continuous on (-3,3)

Also taking the condtion (2) and (3)

$\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{-}}{lim}$ $\sqrt{9-x^2}=0=f\left(3\right)$

$\underset{x\to 3^{+}}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}$ $\sqrt{9-x^2}=0=f\left(3\right)$

Since all the 3 conditions are satisfied we can say f(x) is continuous on [-3,3]