Help!! find absolute extrema of f(x)= (x-1)/(x^2+1) on [0,5]

[kttmaloney]

Hey guys I am needing to find the absolute extrema of the function

f(x)= (x-1)/(x^2+1) ; [0,5]

I got as far as (-x^2-2x+1)/(x^2+1)^2 for the first derivative but don't understand how to get the next couple of steps!


Also need help with this bugger to. I don't do well with fractions so if you would find it in the kindness of your heart to explain in detail I will love you forever.

f(x)= (x^2+18)^2/3 ; [-3,2]

Thanks!

 

[daon2]

Find the critical numbers by setting the derivative you found equal to zero. Note that the denominator you have is always nonzero, so it suffices to set the numerator in your derivative to zero and solve for x.

For all x-values you find, evaluate the original function at these values. Then evaluate the function at the endpoints of your interval.

The smallest of these is the absolute minimum. The largest is the absolute maximum.


For the second function, note that the derivative is, using the chain rule, \(\displaystyle (2/3)\cdot(x^2+18)^{-1/3}\cdot 2x = \dfrac{4x}{3\sqrt[3]{x^2+18}}\)

 

[Steven G]

Hey guys I am needing to find the absolute extrema of the function

f(x)= (x-1)/(x^2+1) ; [0,5]

I got as far as (-x^2-2x+1)/(x^2+1)^2 for the first derivative but don't understand how to get the next couple of steps!


Also need help with this bugger to. I don't do well with fractions so if you would find it in the kindness of your heart to explain in detail I will love you forever.

f(x)= (x^2+18)^2/3 ; [-3,2]

Thanks!

Subtracting 1 from a fraction is simple when you think of it correctly.
a/b - 1 = a/b -b/b =(a-b)/b. So to get the numerator you compute the numerator minus the denominator. The denominator stays the same.

For example: 5/3 - 1 = (5-3)/3 or 2/3.
5/8 - 1 =(5-8)/8 = -3/8.