Mathematics: Need some with this, pic of question

Need some with this, pic of question

Ответ:

ROBIOX3551

10.03.2022

$\boxed{c = 9}.$

Step-by-step explanation:

The limit $\lim\limits_{x\to -3} f(x)$ exists if, and only if both $\lim\limits_{x\to -3^-} f(x)$ and $\lim\limits_{x\to -3^+} f(x)$ exist and $\lim\limits_{x\to -3^-} f(x) = \lim\limits_{x\to -3^+} f(x)$ .

Since f(x) = x^2 - 9 for , we have:

$\lim\limits_{x\to-3^-}f(x) = \lim\limits_{x\to-3^-} (x^2 - 9) = (-3)^2 - 9 = 9-9 = 0.$

On the other hand, f(x) = -x^2+c for x -3 , so we have:

$\lim\limits_{x\to-3^+} f(x) = \lim\limits_{x\to-3^+} (c-x^2) = c-(-3)^2 = c-9.$

Assuming that $c \in \mathbb{R}$ , both one-sided limits exist, which means they must be equal:

$\lim\limits_{x\to -3^-} f(x) = \lim\limits_{x\to -3^+} f(x) \iff 0 = c-9 \iff c = 9.$

So we finally get:

$\boxed{c = 9}.$

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Ответ:

uverworldxdz4999

05.03.2021

Mathematics

0.65 percent

Step-by-step explanation:

because 13/21 was doing it and therefor you divide 13 by 21 and get 0.65

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