Question

Not asking to do my HW but I am so stuck on this question:

Find an open interval about x0 on which the inquality |f(x)-L| < E holds. Then give a value for delta > 0 such that for all x satisfying 0 < |x - x0| < delta the inquality |f(x)-L| < E holds.

F(x) = x^2 L = 3 x0 = Square root of 3 E= .1

I have this formula so far but I don't know how to solve it:

o < |Square root of 3 - 9| |x^2-3| < .1

Answer 1

\[0<|x-3|<\delta \Rightarrow \left|x^2-9\right|<\varepsilon = 0.1\]\[0<|x-3|<\delta \Rightarrow \left|x-3\right|\left|x+3\right|<\varepsilon = 0.1\]
for the moment, let us say \(\delta \leq 0.1\)

\[|x-3|<0.1\Rightarrow -0.1<x-3<0.1\]

add 6 to each part of the continuous inequality \[-0.1+6<x-3+6<0.1+6\\5.9<x+3<6.1\\|x+3|<6.1\]

Answer 2

combine the inequalities
\[|x-3|<\delta\\|x+3|<6.1,\qquad \delta\leq 0.1\]

\[|x-3|\cdot|x+3|<\color{red}{\delta (6.1)}, \qquad \delta \leq 0.1\]

but \[|x-3|\cdot|x+3|=|x^2 - 9| < \color{red}{\varepsilon=0.1}\]

therefore we get \(6.1\delta=\varepsilon=0.1\) by equating the expressions in red and solving \(\delta\)

\[\delta=\frac{1}{61},\qquad \delta \leq 0.1\]

Answer 3

if you choose a different starting value of \(\delta\), say \(\delta\leq 1\), there will be a different solution\[\delta=\frac{0.1}{7},\qquad \delta \leq 1\]

Answer 4

Thanks!!