(a) Show that

Chapter 1, Problem 29

(choose chapter or problem)

(a) Show that

$\left|\left(3 x^{2}+2 x-20\right)-300\right|=|3 x+32|+|x-10|$

(b) Find an upper bound for $|3 x+32|$ if x satisfies $|x-||0|=1$

$\lim _{x \rightarrow 10}\left[3 x^{2}+2 x-20\right]=300$

Suppose that $e=0$. Set $\delta=\min$ (1,_________) and assume that $0=|x-10|=\delta$. Then

$\left|\left(3 x^{2}+2 x-20\right)-300\right|=|3 x+32| \cdot|x-10|$

< _________ $\cdot|x-10|$ < _________$\cdot$_________ $=\epsilon$

Equation Transcription:

$|\left({3{x}^{2}+2x-20}\right)-300|=|3x+32|+|x-10|$

$|3x+32|$

$|x-||0|=1$

$\lim\limits_{{x} \rightarrow {10}}\ [3{x}^{2}+2x-20]=300$

$e=0$

$\delta{}=min$

$0=|x-10|=\delta{}$

$\cdot{}\ |x-10|$

$=\ \epsilon{}$

Text Transcription:

|3x^2+2x-20-300|=|3x+32|+|x-10|

|3x+32|

lim_x rightarrow 10 [3x^2+2x-20]=300

e = 0

delta = min

0=|x-10|=delta

times |x-10|

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