Find the x-coordinates of any relative extrema and inflection point(s) for the function

Question

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x^(1/3) + 3x^(4/3). You must justify your answer using an analysis of f ′(x) and f ′′(x).

Answer 1

6x^(1/3) + 3x^(4/3) =

3x^1/3( 2 + x)

The roots are x=0 and x= -2

Sign chart


Interval Sample test value f(x)

x<-2 -8 36

-2<x<0 -1 -9

x>0 1 9


First derivative:

2x^(-2/3) + 4x^(1/3)

Critical values:
2x^(-2/3) + 4x^(1/3) = 0

2 + 4x = 0 <--- multiplies everything by x^(2/3)

x = -1/2

Critical point is 3*(-1/2)^(1/3) * ( -1/2 +2 ) =
3 (-1/(2^(-1/3))) * ( 3/2) =
(-3/2)* 2^(2/3) * (3/2) =
(-9/4)*2^(2/3)

(-1/2,-3.571652...)

For x < -1/2, like for example x=-1, the first derivative is

2(-1)^(-2/3) + 4(-1)^(1/3) = 2*(-1)^2 + 4(-1) = 2*1 - 4 = 2 - 4 = -2
So the function is decreasing.

For x > -1/3, like for example x=0, the first derivative is 6.
(Note: recall zero to any power is 1. So f'(0) = 2*1 + 4*1 = 6.)
Thus, the function is increasing.


The second derivative:
(-4/3)x^(-5/3) + (4/3)x^(-2/3)

The flex points occur when:
(-4/3)x^(-5/3) + (4/3)x^(-2/3) =0

Multiplies everything by x^(5/3)
(-4/3) + (4/3)X = 0

The flex value is X=1
Again, 1 to any power is 1, so f(1) = 6*1 + 3*1 = 9
The flex point is (1,9)

For x<1, like for example x=0, f''(0) = -4/3
so the function is concave down

for x>1 like for example 3, f''(3) = 4 - 4/3 = 12/3 - 4/3 = 8/3 >0
so the function is concave up

 

You can now get a good graph.

Answer 2

To find the relative extrema, set f'(x) equal to zero and solve for x.

 

2x^-2/3 + 4x^1/3 = 0

 

2x^-2/3(1 + 2x) = 0

 

x = -1/2

 

 

Then you evaluate the derivative around this point to see of derivative changes signs.  It must changes signs to be an extrema.

 

 

To find points of inflection, set f"(x) equal to zero and solve for x.

 

-(4/3)x^-5/3 + (4/3)x^-2/3 = 0

 

(4/3)x^-5/3 (-1 + x^1/3) = 0

 

 

-1 + x^1/3 = 0

 

x^1/3 = 1

 

x = 1

 

 

This is the location of the point of inflection.  Evaluate the second derivative around this point to see if the second derivative changes signs.  It must change signs to be an inflection point.

Answer 3

Take derivative to get f'; take derivative again to get f"

set f' to zero to to find critical values of x, where extrema can occur.

 

you should get f' = 2x^-2/3(x+2) so the critical point is x=-2.

To the left of the crit. pt., f' is negative (f is increasing); to the right of -2, f' is positive so f is decreasing.

That establishes fact that the crit. pt. is a relative maximum of f.

 

To find inflection point, set f" to zero.  IN OTHER WORDS, JUST DO AS THE TEACHER & TEXTBOOK PROCEDURES tell you.