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For what values of x is the linear approximation $\sqrt{x + 3} \simeq \frac74 + \frac x4 $accurate to within 0.5?

My Try

I found this question under llinearization lesson Since this problem does not mention about a point around which it is approximated, I took it as 'a'

$\therefore$ $f(x)=f(a)+(x-a) f'(a)$

$f(x)=\sqrt{a+3} + \frac{x-a}{2\sqrt{x+3}}$

How do I proceed to find 'a' such that the error is within 0.5? Please Help! Thanks in advance.

emil
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1 Answers1

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Hint:

You could try to solve $$\sqrt{x + 3} = \frac74 + \frac x4 +\frac12$$ and $$\sqrt{x + 3} = \frac74 + \frac x4 -\frac12$$ which if you square both sides will give you two quadratics and possibly some spurious solutions to check. Then spot the relevant intervals

Henry
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  • Doesn't it have to be an inequilaity? |f(x)-L(x)|<0.5 – emil Jun 09 '19 at 19:37
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    @emil yes it does, but the solutions will be in the relevant intervals. As a further tip, the first equation does not seem to have real roots – Henry Jun 09 '19 at 19:40
  • Interesting! that's because the function is concave down..Thanks a lot – emil Jun 09 '19 at 19:43
  • @emil: I would say it is because $\left(\frac{x-1}{4}\right)^2 = \left(\frac74 + \frac x4\right)^2-(x+3)$ so $\sqrt{x + 3} \le \frac74 + \frac{x}4 , \forall x \ge -3$ – Henry Jun 09 '19 at 19:49