Will give brainliest if correct

Mathematics

1 answer:

Nuetrik [128]3 years ago

6 0

Answer:

Step-by-step explanation:

$|\angle AEC|=(4x)^o \quad\implies\quad |\angle AED|=180^o-(4x)^o\\\\\\\widehat{AC}=120^o\quad\implies\quad |\angle AOC|=120^o\\\\|\angle AOC|=120^o \quad\implies\quad |\angle ADC|=\frac12\cdot120^o=60^o\\\\\\\widehat{BD}=(2x)^o\quad\implies\quad |\angle BOD|=(2x)^o\\\\|\angle BOD|=(2x)^o \quad\implies\quad |\angle ADC|=\frac12\cdot(2x)^o=x^o\\\\\\ \Delta ADE:\\\\180^o-(4x)^o+60^o+x^o=180^o\\\\-(3x)^o=-60^o\\\\x=20^o\\\\\\|\angle AED|=180^o-4\cdot20^o=100^o$

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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:

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