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shwanky
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I'm a little nervous about a test I have on Thursday and I was wondering if this is adequate for a proof of the following equation.
Show that f(x) is continuous at a=4
[tex]f(x) = x^2 + \sqrt{7-x}[/tex]
For f(x) to be continuous at a = 4
a) f(a) must be defined
b) [tex]\lim_{x \to a} f(x)[/tex] must exist
c) [tex]\lim_{x \to a} f(x) = f(a)[/tex]
Proof
1. [tex]f(a) = f(4) = 4^2 + \sqrt{7-4} = 16 + \sqrt{3}[/tex]
Therefore, f(a) is defined at a=4.
2. [tex]\lim_{x \to a} f(x) = \lim_{x \to 4} x^2 + \sqrt{7-x}[/tex]
[tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]
[tex]\lim_{x \to 4^-} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]
Since [tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3} = \lim_{x \to 4^-} x^2 + \sqrt{7-x} = \lim_{x \to 4} f(x)[/tex] exists.
[tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]
3. By steps 1 and 2 we know that f(4) = [tex] 16 + \sqrt{3}[/tex] and [tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex].
Since [tex]\lim_{x \to 4} f(x) = 16 + \sqrt{3} = f(4)[/tex], [tex]f(x) = x^2 + \sqrt{7-x}[/tex] is continuous at a = 4.
Homework Statement
Show that f(x) is continuous at a=4
Homework Equations
[tex]f(x) = x^2 + \sqrt{7-x}[/tex]
The Attempt at a Solution
For f(x) to be continuous at a = 4
a) f(a) must be defined
b) [tex]\lim_{x \to a} f(x)[/tex] must exist
c) [tex]\lim_{x \to a} f(x) = f(a)[/tex]
Proof
1. [tex]f(a) = f(4) = 4^2 + \sqrt{7-4} = 16 + \sqrt{3}[/tex]
Therefore, f(a) is defined at a=4.
2. [tex]\lim_{x \to a} f(x) = \lim_{x \to 4} x^2 + \sqrt{7-x}[/tex]
[tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]
[tex]\lim_{x \to 4^-} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]
Since [tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3} = \lim_{x \to 4^-} x^2 + \sqrt{7-x} = \lim_{x \to 4} f(x)[/tex] exists.
[tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]
3. By steps 1 and 2 we know that f(4) = [tex] 16 + \sqrt{3}[/tex] and [tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex].
Since [tex]\lim_{x \to 4} f(x) = 16 + \sqrt{3} = f(4)[/tex], [tex]f(x) = x^2 + \sqrt{7-x}[/tex] is continuous at a = 4.
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