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Lelechka [254]
3 years ago
10

Complete the table above.

Mathematics
1 answer:
4vir4ik [10]3 years ago
4 0

\bf \begin{array}{|c|c|c|ll}
\cline{1-3}
x&y=3x^2&y=3^x\\
\cline{1-3}&&\\
0&\stackrel{3(0)^2}{0}&\stackrel{3^0}{1}\\&&\\
1&\stackrel{3(1)^2}{3}&\stackrel{3^1}{3}\\&&\\
2&\stackrel{3(2)^2}{12}&\stackrel{3^2}{9}\\&&\\
\cline{1-3}
\end{array}

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Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x &gt; 0 [−4, 5]The f
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It is continuous since \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)

Step-by-step explanation:

We are given that the function is defined as follows f(x) = 9-x, x\leq 0 and f(x) = 9+12x, x>0 and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity)  x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all \mathbb{R}. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

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\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

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