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Blizzard [7]
2 years ago
12

If limx→3f(x)=7, which of the following must be true? I. f is continuous at x = 3 II. f is differentiable at x = 3

Mathematics
1 answer:
Viktor [21]2 years ago
3 0

Without knowing anything else about f(x), neither of these need be true.

Suppose

f(x)=|x-3|+7=\begin{cases}10-x&\text{for }x3\\0&\text{for }x=3\end{cases}

Then (I) isn't true because, while the limit exists as x\to3 and is equal to 7, we have f(3)=0\neq7, so f(x) is not continuous there.

(II) also true because f(x) is not differentiable at x=3; that is,

\displaystyle\lim_{x\to3^-}f'(x)=\lim_{x\to3}(10-x)'=\lim_{x\to3}(-1)=-1

but

\displaystyle\lim_{x\to3^+}f'(x)=\lim_{x\to3}(x+4)'=\lim_{x\to3}1=1

which means the derivative does not exist at x=3.

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Explanation:

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2 years ago
0.2x + 0.3y = 1.3.<br>0.4r + 0.5y = 2.3. solve by using substitution ​
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Answer:

x=2, y=3.

Step-by-step explanation:

0.2x + 0.3y = 1.3

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Substituting in second equation:

0.4x + 0.5(1.3 - 0.2x) / 0.3 = 2.3

Multiply through by 0.3:

0.12x + 0.5(1.3 - 0.2x) =  0.69

0.12x + 0.65 - 0.1x = 0.69

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Substituting for x in equation 1:

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Check in equation 2:

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