- #1

srvs

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At least I think it's via linearization.

Let

[itex]f(x) = \tan (x^2) - 1[/itex]

and

[itex]g(x) = \frac{\ln((x+1)^3)}{3}[/itex]

Find the smallest positive and negative intersection with a relative error of less than 0.001.

I don't know. You can linearize one or both, yeah, but you don't have any analytical value to compare it with, how would go determine the relative error? Can anyone put me on the right track?

Let

[itex]f(x) = \tan (x^2) - 1[/itex]

and

[itex]g(x) = \frac{\ln((x+1)^3)}{3}[/itex]

Find the smallest positive and negative intersection with a relative error of less than 0.001.

I don't know. You can linearize one or both, yeah, but you don't have any analytical value to compare it with, how would go determine the relative error? Can anyone put me on the right track?

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