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February 27, 2015

February 27, 2015

Posted by **Bob** on Thursday, January 8, 2009 at 7:36pm.

f(x)=2ln(x^2+3)-x with domain -3 is less than or equal to x which is less than or equal to 5

a) Find the x-coordinate of each relative maximum point and each relative minimum point of f. Justify your answer.

b) Find the x-coordinate of each inflection point of f.

c) Find the absolute maximum value of

f(x).

- Calc -
**Reiny**, Thursday, January 8, 2009 at 9:13pmf'(x) = 2(2x)/(x^2 + 1) - 1

= 0 for max/min

for this I got x = 2 ± √5

These lie within your domain so

You will have to find

f(-3), f(2+√5)), f(2-√5) and f(5) to see which is the maximum

for f''(x) I got (4(x^2 + 1) - 4x(2x))/(x^2 + 2)^2

setting this equal to zero, I got x = ± √2

sub back in the original to find the two inflection points.

f(-3)

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