Question

Tough calculus question I have been struggling with, please help me

Answer 1

You have the following function:

`f(x)=9x^{(1)/(3)}+(9)/(2)x^{(4)/(3)}`

The first derivative of the previous function is:

`\begin{gathered} f^(\prime)(x)=(1)/(3)\cdot9x^{(1)/(3)-1}+(4)/(3)\cdot(9)/(2)x^{(4)/(3)-1} \\ f^(\prime)(x)=3x^{-(2)/(3)}+6x^{(1)/(3)} \end{gathered}`

next, equal the previous expression to 0 and solve for x:

`\begin{gathered} 3x^{-(2)/(3)}+6x^{(1)/(3)}=0 \\ (x^{-(2)/(3)})^3=(-2x^{(1)/(3)})^3 \\ x^(-2)=-8x \\ (1)/(x^2)=-8x^{} \\ -(1)/(8)=x^3 \\ x=-(1)/(2) \end{gathered}`

Then, for x=1/2 there is a critical point.

To determine if the critical point is a maximum or a minmum, derivate f'(x) and replace x=1/2=0.5:

`\begin{gathered} f^(\doubleprime)(x)=-(2)/(3)\cdot3x^{-(5)/(3)}+(1)/(3)\cdot6x^{-(2)/(3)} \\ f^(\doubleprime)(0.5)=-2(-0.5)^{-(5)/(3)}+2(-0.5)^{-(2)/(3)}\approx9.52>0 \end{gathered}`

Now, by considering the second derivative test, you have that if f(xo)>0, then, there is a local minimum at xo.

In this case, there is a local minimum at x = -0.5.

The y-coordinate is found by replacing x=-0.5 into f(x):

`f(-0.5)=9(-0.5)^{(1)/(3)}+(9)/(2)(-0.5)^{(4)/(3)}=-5.357`

Then, the relative extrema point is (-0.5 , -5.357)

The inflection point is found by making the expression for f'' = 0 and solving for x:

`\begin{gathered} -2x^{-(5)/(3)}+2x^{-(2)/(3)}=0 \\ -x^{-(5)/(3)}+x^{-(2)/(3)}=0 \\ (x^{-(5)/(3)})^3=(x^{-(2)/(3)})^3 \\ x^(-5)=x^(-2) \\ x^(-5)\cdot x^2=1 \\ x^(-3)=1 \\ x^3=1 \end{gathered}`

Then, for x = 1 there is an inflection point.

By replacing x=1 into f(x):

`f(1)=9(1)^{(1)/(3)}+(9)/(2)(1)^{(4)/(3)}=9+(9)/(2)=13.5`

Hence, the inflection point is (1 , 13.5)