AP calculus, calculus

Question

Let f be the function given by f(x)=ln|x/1+x^2|. At what values of x does f have a relative maximum or a relative minimum? For each such x, use the first derivative test to determine whether f(x) is a relative maximum or a relative minimum.

Answer 1

f(x)=ln |x/1+x^2| , because we have ln function , abs[ x/1+x^2] >0, since in this case the denominator always positive , therefore x can be positive and negative but not equal zero.

Domain of x : (-∞,0) U (0,∞ )

To find the relative max or min , we can drive the function ,and to make it easy ,write the function in this form,

f(x) = ln x - ln ( 1+ x^2 )

f ' (x) = 1/x - 2 x / ( 1+x^2)

= (1+ x^2 - 2x^2 ) / x (1+x^2 )

= ( 1-x^2) / x(1+ x^2)

1- find zeroes : 1- x^2 =0 , x^2 = 1, either x = +1 or x= -1

2- find undefined : x(1+ x^2) = 0 , either x=0 ( must be neglected, out of domain)

or 1+ x^2 =0 , x^2 = -1 ( neglect , this gives not real values)

so the possible relative extrema value is at x=1 , x=-1

we have to see if at x=1,x= -1 we have relative max or min,

take value less than x=1 for example x=1/2 , plug this value in f'

then take value greater than x=1 for example x=2 , plug this value in f'

f' (1/2) = ( 1-1/4) / [1/2(1+1/ 4)] = (3/4) /( 5/8) = + 6/5 positive value

f'( 2) = (1-4) / 2(1+4) = -3/10 negative value

we see that f'( slope) changes from + to - which means that we have relative max at ( x=1 , f(1) = ln 1/2 )

Now take value less than x=-1 for example x=-2 , plug this value in f'

then take value greater than x=-1 for example x=-1/2 , plug this value in f'

f' (-2) = ( 1-4) /[ (-2)(1+ 4)] = ( -3) /(-10) =+ 3/10 positive value

f'(-1/2) = (1-1/4) /[(-1/ 2)(1+1/4)] =(3/4) / (-5/8) = - 6/5 negative value

we see that f'( slope) changes from + to - which means that we have another relative max at ( x= -1 , f(-1) = ln 1/2 )

Hint: there is another way to check if we have relative max or min by using 2nd derivative test :

f ' (x) = 1/x - 2 x / ( 1+x^2)

f'' (x) = -2/ x^2 - [ 2(1+x^2)- 2x*2x]/ ( 1+x^2) ^2

= -2/ x^2 -( 2-2x^2)/ ( 1+x^2) ^2

f''(1) = -2 negative value , so the function is concave down, and that's mean we have relative max at (1, ln 1/2)

f''(-1) =-2 negative value, so the function is concave down, and that's mean we have relative max at (-1, ln 1/2)

Answer 2

Answer: The function f(x)=ln[x/(1+x²)] has a relative maximum at x = 1 (or (1, ln(1/2)))

Explanation: Before we do anything, we have to know what values of x are possible answers. In other words, we need to be aware of any domain restrictions. ln(u) only has values for any u > 0 (where u can be any function of x) so for our parent function, x/(1+x²) > 0. By multiplying both sides by (1+x²), we derive that for this problem x > 0 so any extrema we find must have an x value larger than 0.

To solve the problem, we first take the derivative of f. The derivative of ln(u) is 1/u (du comes in because of the chain rule) so the first part of the derivative is (1/(x/(1+x²))) OR (1+x²)/x. Then, since there is a nested function of x within ln(u), we must apply the chain rule and multiply the derivative we already found (1+x²)/x by the derivative of the nested function x/(1+x²). The derivative of x/(1+x²) is (1-x²)/(1+x²)². The (1+x²) from the first part of the derivative cancels out with the (1+x²)² in the second part of the derivative, making the full derivative of f(x) equal to ((1+x²)/x)((1-x²)/(1+x²)²) or f'(x) = (1-x²)/(x(1+x²)).

To apply the First Derivative Test, we must find values of x where f'(x) is either 0 or undefined. f'(x) will equal 0 when x =1 or x =-1 since the numerator (1-x²) will equal zero and the denominator will equal either 1/2 or -1/2. However, since our domain (set of possible x values) is restricted to positive values of x only (x>0), x = -1 cannot be one of the possible answers. For finding values where x is undefined, we could try the value x = 0 where the denominator will equal zero so f'(0) = undefined. However, just like x = -1, x = 0 is not within the set of possible answers to the problem since x > 0. Therefore, the only possible extrema can be located at x = 1.

The second part of the First Derivative Test is used to determine whether x =1 is a maximum or a minimum. We know that if the slope of f(x) or value of the First Derivative switches signs from negative to positive (therefore intersecting the y-axis), the corresponding x-value is a minimum but if the slope/ value of first derivative switches from positive to negative, the corresponding x-value is a maximum. Therefore, if we know what the sign of f'(x) to the left of and right x =1, we can determine whether x is a min or max. The slope to the left of x = 1 can be tested by plugging in x = 1/2 to f'(x), where f'(1/2) = 1.2 and is positive. The slope to the right of x = 1 can be tested by plugging in x = 2 to f'(x) where f'(2) = -0.3 and is negative. Since the slope of f(x)/ value of f'(x) changes from positive to negative, we can conclude that x = 1 is a maximum.