Is (-2,8), (-1,5), (0,4), (1,5), and (2,8) a function?

Mathematics

1 answer:

Dimas [21]2 years ago

7 0

Answer:

$y = {x}^{2} + 4$

Step-by-step explanation:

First you can notice the y-intercept is 4. Then notice how the y value increases exponentially based on the x valie, making it non-linear, rather a porabla.

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write in scientific notation 1. 0.0000364 2. 0.00751 3. 0.10005 4. 1,094 5. 0.00000099 6. 0.04101 7. 10,500 8. 8,9000

solmaris [256]

$1.\\0\underbrace{.00003}_{5\to}54=3.54\times10^{-5}\\2.\\0\underbrace{.007}_{3\to}51=7.51\times10^{-3}\\3.\\0\underbrace{.1}_{1\to}0005=1.0005\times10^{-1}\\4.\\1\underbrace{,094}_{\leftarrow3}=1.094\times10^3\\5.\\0\underbrace{.0000009}_{7\to}9=9.9\times10^{-7}$
$6.\\0\underbrace{.04}_{2\to}101=4.101\times10^{-2}\\7.\\1\underbrace{0,500}_{\leftarrow4}=1.05\times10^4\\8.\\8\underbrace{9000}_{\leftarrow4}=8.9\times10^4$

3 0

2 years ago

For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course

pav-90 [236]

By definition of absolute value, you have

$f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1$

or more simply,

$f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x$

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

$f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x$

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -1 for the derivative do not match:

$\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1$