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13 votes
13 votes
Find the values of x for which the function is continuous.

f(x) =

−3x + 7 if x < 0
x^2 + 7 if x ≥ 0

a.x ≠ 0

b.x ≥ 0


c.x ≥ root square 7
d.all real numbers

e.x ≠ 7

User Arturgspb
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2.7k points

1 Answer

21 votes
21 votes

It looks like the given function is


f(x) = \begin{cases}-3x + 7 &amp; \text{if }x < 0 \\ x^2+7 &amp; \text{if }x \ge0\end{cases}

The two pieces of f(x) are continuous since they are polynomials, so we only need to worry about the point at which they meet, x = 0. f(x) is continuous there if


\displaystyle \lim_(x\to0^-) f(x) = \lim_(x\to0^+) f(x) = f(0) = 7

To the left of x = 0, we have x < 0, so f(x) = -3x + 7 :


\displaystyle \lim_(x\to0^-) f(x) = \lim_(x\to0) (-3x + 7) = 7

To the right of x = 0, we have x > 0 and f(x) = x² + 7 :


\displaystyle \lim_(x\to0^+) f(x) = \lim_(x\to0) (x^2 + 7) = 7

So f(x) is continuous at x = 0, and hence continuous for all real numbers.

User HDW Production
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