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

So your epsilon is 0.2, and your limit is 3\[\left| x - 2 \right| < \delta \rightarrow \left| \sqrt[5]{x-2 }+ 3 - 3\right| < 0.2\]

Answer 2

\[\left| \sqrt[5]{x-2} \right| < 0.2\]

Answer 3

\[\left| x-2 \right| < 0.2^{5}\]

Answer 4

\[\left| x-2 \right| < .00032\]

Answer 5

So, delta = .00032

Answer 6

That is, when x is within .00032 units of 2, f(x) is within 0.2 units of 3

Answer 7

I hope that helps!