Calculus Question

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pennyhaehcf

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Use the given graph of
f(x) = x^2
to find a number δ such that
abs(x^2 minus 2) less than (1 over 2)
whenever
abs(x minus sqrt(2)) less than delta
.


A curve, 2 vertical solid lines, 2 horizontal solid lines, 1 horizontal dashed line, and 1 vertical dashed line are graphed on the x y coordinate plane. The curve enters the top left of the viewing window in the second quadrant, goes down and to the right, passes through the origin, goes up and to the right, and exits the top right of the viewing window. The first horizontal solid line starts at a point (0, 1.5), goes horizontally to the right, and ends at a point on the curve whose x-coordinate is unknown and is labeled question mark and whose y-coordinate is 1.5. Then, a vertical solid line starts at this point, goes down vertically, and ends at a point on the x-axis whose x-coordinate is unknown and is labeled question mark and whose y-coordinate is 0. The dashed horizontal line starts at the point (0, 2), goes horizontally to the right, and ends at the approximate point (2,1.41) on the curve. The dashed vertical line starts at this point, goes down vertically, and ends at the approximate point (1.41, 0). The other solid horizontal line starts at the point (0, 2.5), goes horizontally to the right, and ends at a point on the curve whose x-coordinate is unknown and is labeled question mark and whose y-coordinate is 2.5. The solid vertical line starts from this point, goes down vertically, and ends at a point on the x-axis whose x-coordinate is unknown and labeled question mark and whose y-coordinate is 0.

It'd be helpful if you show the steps, I'm stuck
 
pennyhaehcf said:
Use the given graph of
f(x) = x^2
to find a number δ such that
abs(x^2 minus 2) less than (1 over 2)
whenever
abs(x minus sqrt(2)) less than delta
.


A curve, 2 vertical solid lines, 2 horizontal solid lines, 1 horizontal dashed line, and 1 vertical dashed line are graphed on the x y coordinate plane. The curve enters the top left of the viewing window in the second quadrant, goes down and to the right, passes through the origin, goes up and to the right, and exits the top right of the viewing window. The first horizontal solid line starts at a point (0, 1.5), goes horizontally to the right, and ends at a point on the curve whose x-coordinate is unknown and is labeled question mark and whose y-coordinate is 1.5. Then, a vertical solid line starts at this point, goes down vertically, and ends at a point on the x-axis whose x-coordinate is unknown and is labeled question mark and whose y-coordinate is 0. The dashed horizontal line starts at the point (0, 2), goes horizontally to the right, and ends at the approximate point (2,1.41) on the curve. The dashed vertical line starts at this point, goes down vertically, and ends at the approximate point (1.41, 0). The other solid horizontal line starts at the point (0, 2.5), goes horizontally to the right, and ends at a point on the curve whose x-coordinate is unknown and is labeled question mark and whose y-coordinate is 2.5. The solid vertical line starts from this point, goes down vertically, and ends at a point on the x-axis whose x-coordinate is unknown and labeled question mark and whose y-coordinate is 0.

It'd be helpful if you show the steps, I'm stuck
Click to expand...
Please calculate the domain of:

\(\displaystyle \left| x^2 - 2\right| \lt \frac{1}{2}\)

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.
 
pennyhaehcf said:
Use the given graph of
f(x) = x^2
to find a number δ such that
abs(x^2 minus 2) less than (1 over 2)
whenever
abs(x minus sqrt(2)) less than delta
.
It'd be helpful if you show the steps, I'm stuck.
Click to expand...
Start with \(|x-\sqrt2|<1\) then we know that \(-1<x-\sqrt2<1\text{ or }2\sqrt2-1<x+\sqrt2<1+2\sqrt2\).
Now \(\max\{1+2\sqrt2,-1+2\sqrt2\}\le 4\) so that\(|x+\sqrt2|<4\) if \(|x-\sqrt2|<1\)
Note that \(|x^2-2|=|x+\sqrt2|\cdot|x-\sqrt2|<4\cdot|x-\sqrt2|\).
Can you finish?
 
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