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

1

Explanation:

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

2x-2/3 + 4x1/3 = 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 + x1/3) = 0

-1 + x1/3 = 0

x1/3 = 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 2

Applying our power rule gets us our first derivative,


`\rm f'(x)=6\frac13x^(-2/3)+3\cdot\frac43x^(1/3)`

simplifying a little bit,


`\rm f'(x)=2x^(-2/3)+4x^(1/3)`

looking for critical points,


`\rm 0=2x^(-2/3)+4x^(1/3)`

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.


`0=2x^(-2/3)\left(1+2x\right)`

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,


`\rm 0=2x^(-2/3)\qquad\qquad\qquad 0=(1+2x)`

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,


`\rm f'(x)=2x^(-2/3)+4x^(1/3)`

and take our second derivative.


`\rm f''(x)=-\frac43x^(-5/3)+\frac43x^(-2/3)`

Looking for inflection points,


`\rm 0=-\frac43x^(-5/3)+\frac43x^(-2/3)`

Again, pulling out the smaller power of x, and fractional part,


`\rm 0=-\frac43x^(-5/3)\left(1-x\right)`

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.