Question

For f(x)= (x-2)^1/5 + 3, calculate the maximum value that delta can be so that f(x) is within 0.2 unit of 3 when x is within delta units of 2.

Answer 1

If \(|f(x)-3|\leq0.2\), then \(|(x-2)^{1/5}+3-3|\leq0.2\). If you simplify what's inside the absolute value bars and use the property of absolute value where \(|b^a|=|b|^a\), then you should see the answer.

Answer 2

why did you put in absolute values signs? and why is there only one side to the inequality.? we've been practiving using 2 sides then choosing the smaller delta or something like that

Answer 3

@blockcolder could you do \[2.8 \pm (x-2)^{1/5} + 3 < 3.2\] ?

Answer 4

oops the first sign is supposed to be a less than

Answer 5

Yes. The absolute value notation is simply a shorthand. If you use the fact that
\[|x-a|\leq b \Leftrightarrow -b\leq x-a \leq b\]
Then you can rewrite my absolute value statement as
\[\begin{align}
|(x-2)^{1/5}+3-3|\leq0.2 &\Leftrightarrow -0.2 \leq (x-2)^{1/5}+3-3\leq 0.2\\
&\Leftrightarrow 2.8 \leq (x-2)^{1/5}+3\leq 3.2
\end{align}\]

Answer 6

I thought this absolute value notation was the more commonly taught form.

Answer 7

maybe we haven't gotten that far yet or something...then you just solve it algebraically, correct? and subtract the smaller delta from the x?

Answer 8

@blockcolder

Answer 9

Lemme just check.
The limit definition goes something like this, right?
\[|x-a|\leq\delta \Rightarrow |f(x)-L|\leq\epsilon\]

Answer 10

i believe so

Answer 11

Then my absolute value notation is convenient, because if you simplify my expression, then you have
\[|(x-2)^{1/5}|\leq0.2 \Rightarrow |x-2|^{1/5}\leq 0.2 \Rightarrow |x-2|\leq 0.2^5\]
So that the biggest delta you can use is 0.2^5.

Answer 12

my books answer is .00032

Answer 13

Yeah, and 0.2^5=0.00032 (check with your calculator)

Answer 14

oh ok thanks!