<img src="https://tex.z-dn.net/?f=%5Clongmapsto%5Csqrt%7B64%7D" id="TexFormula1" title="\longmapsto\sqrt{64}" alt="\longmapsto\s

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liubo4ka [24]

2 years ago

qrt{64}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

natulia [17]2 years ago

7 0

$\sqrt{64} = 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$\sqrt{64} = 2 \times 2 \times 2$

$\sqrt{64} = 8$

<h3># Answered by Cutest Ghost..</h3>

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Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 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

$\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$ (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

$\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9$ . 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.