f'(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)
Let f be the function given by
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).
1 answer