What are the local extrema, if any, of #f (x) =(x^3 + 2x^2)/(3 - 5x)#?

1 Answer
Nov 22, 2016

Answer:

Local Extrema:

#x~~-1.15#

#x=0#

#x~~1.05#

Explanation:

Find the derivative #f'(x)#

Set #f'(x)=0#

These are your critical values and potential local extrema.

Draw a number line with these values.

Plug in values within each interval;

if #f'(x) >0#, the function is increasing.

if #f'(x) <0#, the function is decreasing.

When the function changes from negative to positive and is continuous at that point, there is a local minimum; and vice versa.

#f'(x)=[(3x^2+4x)(3-5x)-(-5)(x^3+2x^2)]/(3-5x)^2#

#f'(x)=[9x^2-15x^3+12x-20x^2+5x^3+10x^2]/(3-5x)^2#

#f'(x)=(-10x^3-x^2+12x)/(3-5x)^2#

#f'(x)=[-x(10x^2+x-12)]/(3-5x)^2#

Critical values:

#x=0#

#x=(sqrt(481)-1)/20~~1.05#
#x=-(sqrt(481)+1)/20~~-1.15#

#x!=3/5#

<------#(-1.15)#------#(0)#-----#(3/5)#-----#(1.05)#------>

Plug in values between these intervals:

You will get a:
Positive value on #(-oo, -1.15)#
Negative on #(-1.15, 0)#
Positive on #(0, 3/5) #
Positive on #(3/5, 1.05)#
Negative on #(1.05, oo)#

#:.# Your local maximums will be when:

#x=-1.15 and x=1.05#

Your local minimum will be when:

#x=0#