You have the following function:
[tex]f(x)=9x^{\frac{1}{3}}+\frac{9}{2}x^{\frac{4}{3}}[/tex]The first derivative of the previous function is:
[tex]\begin{gathered} f^{\prime}(x)=\frac{1}{3}\cdot9x^{\frac{1}{3}-1}+\frac{4}{3}\cdot\frac{9}{2}x^{\frac{4}{3}-1} \\ f^{\prime}(x)=3x^{-\frac{2}{3}}+6x^{\frac{1}{3}} \end{gathered}[/tex]next, equal the previous expression to 0 and solve for x:
[tex]\begin{gathered} 3x^{-\frac{2}{3}}+6x^{\frac{1}{3}}=0 \\ (x^{-\frac{2}{3}})^3=(-2x^{\frac{1}{3}})^3 \\ x^{-2}=-8x \\ \frac{1}{x^2}=-8x^{} \\ -\frac{1}{8}=x^3 \\ x=-\frac{1}{2} \end{gathered}[/tex]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:
[tex]\begin{gathered} f^{\doubleprime}(x)=-\frac{2}{3}\cdot3x^{-\frac{5}{3}}+\frac{1}{3}\cdot6x^{-\frac{2}{3}} \\ f^{\doubleprime}(0.5)=-2(-0.5)^{-\frac{5}{3}}+2(-0.5)^{-\frac{2}{3}}\approx9.52>0 \end{gathered}[/tex]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):
[tex]f(-0.5)=9(-0.5)^{\frac{1}{3}}+\frac{9}{2}(-0.5)^{\frac{4}{3}}=-5.357[/tex]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:
[tex]\begin{gathered} -2x^{-\frac{5}{3}}+2x^{-\frac{2}{3}}=0 \\ -x^{-\frac{5}{3}}+x^{-\frac{2}{3}}=0 \\ (x^{-\frac{5}{3}})^3=(x^{-\frac{2}{3}})^3 \\ x^{-5}=x^{-2} \\ x^{-5}\cdot x^2=1 \\ x^{-3}=1 \\ x^3=1 \end{gathered}[/tex]Then, for x = 1 there is an inflection point.
By replacing x=1 into f(x):
[tex]f(1)=9(1)^{\frac{1}{3}}+\frac{9}{2}(1)^{\frac{4}{3}}=9+\frac{9}{2}=13.5[/tex]Hence, the inflection point is (1 , 13.5)
