How do you find the local maximum and minimum values of #f(x) = x / (x^2 + 81)# using both the First and Second Derivative Tests?

Question

Elijah Abram · Accepted Answer

Minimum #(-9,-1/18)# , Maximum #(9,1/18)#

#f(x)=x/(x^2+81)#

#D_f=RR#

#f'(x)=(x^2+81-x(2x))/(x^2+81)^2# #=#

#(-x^2+81)/(x^2+81)^2# #=#

#-(x^2-81)/(x^2+81)^2#

#f'(x)=0 ##<=> x^2-81=0 <=> x^2=81 <=> (x=9 or x=-9)#

Let's plug in #f'# a value , #x_1in##(-9,9)# . For example, for #x_1=0#
we get

#f'(0)=-(-81)/81^2=81/81^2>0#

Let's plug in #f'# a value which is #>9# . For example, for #x_2=10#
we get

#f'(10)=-(100-81)/(100+81)^2=-19/181^2<0#

Let's plug in #f'# a value which is #<-9# . For example, for #x_3=-10#
we get

#f'(-10)=-(100-81)/(100+81)^2=-19/181^2<0#

As a result we have:

#f# continuous in #(-oo,-9]# and #f'(x)<0# for #x##in##(-oo,-9)#
so #f# is strictly decreasing in #(-oo,-9]#
#f# continuous in #[-9,9]# and #f'(x)>0# for #x##in##(-9,-9)#
so #f# is strictly increasing in #[-9,9]#
#f# continuous in #[9,+oo)# and #f'(x)<0# for #x##in##(9,+oo)#
so #f# is strictly decreasing in #[9,+oo)#
#f# is decreasing in #(-oo,-9]# and increasing in #[-9,9]#
therefore #f# has a local minimum at #x=-9#
#f# is increasing in #[-9,9]# and decreasing in #[9,+oo)#
therefore #f# has a local maximum at #x=9#

Ezra Adams · Answer

undefined To find the local maximum and minimum values of $ f(x) = \frac{x}{x^2 + 81} $ using the First and Second Derivative Tests, follow these steps:

1. **First Derivative Test:**
   - Find the critical points by setting the derivative equal to zero and solving for $ x $.
   - Calculate the derivative of $ f(x) $, $ f'(x) $.
   - Set $ f'(x) = 0 $ and solve for $ x $.
   - Evaluate $ f'(x) $ for each critical point to determine whether the function is increasing or decreasing around those points.

2. **Second Derivative Test:**
   - Find the second derivative of $ f(x) $, $ f''(x) $.
   - Evaluate $ f''(x) $ at each critical point found in step 1.
   - If $ f''(x) > 0 $ at a critical point, then the function has a local minimum at that point.
   - If $ f''(x) < 0 $ at a critical point, then the function has a local maximum at that point.

3. **Conclusion:**
   - Compare the results obtained from the First and Second Derivative Tests to identify local maximum and minimum values.

By following these steps, you can determine the local maximum and minimum values of the function $ f(x) = \frac{x}{x^2 + 81} $ using both the First and Second Derivative Tests.