How do you find the relative extrema for #f(x)=(9x^(2)+1)/x#?

Question

Charles Arocha · Accepted Answer

Relative maxima at x= #-1/3# and minima at x= #1/3)

#(df)/dx= 9-1/x^2#. Equating it to 0 , #x=+-1/3#. f'(x) does not exist at x=0 hence the critical points are #-1/3, 0. and +1/3#. For using first derivative test, test the increasing/decreasing behaviour in intervals #(-oo,-1/3), and (1/3, oo)#. If f '(x) is positive, then f(x) is increasing and if it is negative, then f(x) is decreasing. Take up any test value say -1 in #(-oo,-1/3), #-1/6 in #(-1/3, 0) , #1/6 in (0, 1/3) and +1 in # (1/3, oo)# f ' (x) is positive in #(-oo, -1/3)#, negative in #(-1/3,0)#, negative in #(0, 1/3) #and positive in #(1/3, oo)# Conclusion is there is a relative maxima at x=# -1/3 # ( function changes from increasing to decreasing) and relative minima at x= #1/3# (function changes from decreasing to increasing)

Elijah Abram · Answer

undefined To find the relative extrema for $ f(x) = \frac{9x^2 + 1}{x} $, follow these steps:

1. Find the first derivative of the function $ f(x) $.
2. Set the first derivative equal to zero and solve for $ x $.
3. Determine the corresponding $ y $-values for each critical point found in step 2.
4. Check the nature of each critical point by using the second derivative test or by analyzing the sign changes around each critical point.
5. Classify each critical point as a relative maximum, relative minimum, or neither based on the nature of the critical point.

Let's find the solution:

1. $ f'(x) = \frac{d}{dx}\left(\frac{9x^2 + 1}{x}\right) $
2. Set $ f'(x) = 0 $ and solve for $ x $.
3. Determine the corresponding $ y $-values for each critical point.
4. Analyze the nature of each critical point.
5. Classify each critical point as relative maximum, relative minimum, or neither.

Evelyn Agard · Answer

undefined To find the relative extrema for $ f(x) = \frac{9x^2 + 1}{x} $, we first find the critical points by taking the derivative of the function and setting it equal to zero. Then, we check the second derivative to determine whether each critical point corresponds to a relative maximum, minimum, or neither.

1. Find the derivative of the function:
   $ f(x) = \frac{9x^2 + 1}{x} $
   $ f'(x) = \frac{d}{dx} \left( \frac{9x^2 + 1}{x} \right) $
   $ f'(x) = \frac{(18x)(x) - (9x^2 + 1)(1)}{x^2} $
   $ f'(x) = \frac{18x^2 - 9x^2 - 1}{x^2} $
   $ f'(x) = \frac{9x^2 - 1}{x^2} $

2. Set the derivative equal to zero and solve for $ x $ to find critical points:
   $ f'(x) = \frac{9x^2 - 1}{x^2} = 0 $
   $ 9x^2 - 1 = 0 $
   $ 9x^2 = 1 $
   $ x^2 = \frac{1}{9} $
   $ x = \pm \frac{1}{3} $

3. Check the second derivative at each critical point to determine the nature of the relative extrema:
   Second derivative:
   $ f''(x) = \frac{d^2}{dx^2} \left( \frac{9x^2 - 1}{x^2} \right) $
   $ f''(x) = \frac{d}{dx} \left( \frac{9x^2 - 1}{x^2} \right) $
   $ f''(x) = \frac{(18x)(x^2) - (9x^2 - 1)(2x)}{x^4} $
   $ f''(x) = \frac{18x^3 - 18x^3 + 2x}{x^3} $
   $ f''(x) = \frac{2x}{x^3} $
   $ f''(x) = \frac{2}{x^2} $

At $ x = \frac{1}{3} $:
   $ f''\left(\frac{1}{3}\right) = \frac{2}{\left(\frac{1}{3}\right)^2} = 18 $ (positive, so relative minimum)

At $ x = -\frac{1}{3} $:
   $ f''\left(-\frac{1}{3}\right) = \frac{2}{\left(-\frac{1}{3}\right)^2} = 18 $ (positive, so relative minimum)

Therefore, the relative extrema for $ f(x) = \frac{9x^2 + 1}{x} $ are both relative minima at $ x = \frac{1}{3} $ and $ x = -\frac{1}{3} $.