this is section 3.3 problem 20: for the function f(x)=3x^4 - 4x^3, determine its inflection points. use…

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, for $f(x)=3x^{4}-4x^{3}$, we have $f'(x)=12x^{3}-12x^{2}$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=36x^{2}-24x = 12x(3x - 2)$.

Step3: Set the second - derivative equal to zero

Set $f''(x)=0$, so $12x(3x - 2)=0$. This gives two solutions: $x = 0$ and $x=\frac{2}{3}$ by setting each factor equal to zero ($12x=0$ gives $x = 0$ and $3x - 2=0$ gives $x=\frac{2}{3}$).

Step4: Find the y - values

When $x = 0$, $y=f(0)=3(0)^{4}-4(0)^{3}=0$. When $x=\frac{2}{3}$, $y=f(\frac{2}{3})=3(\frac{2}{3})^{4}-4(\frac{2}{3})^{3}=3\times\frac{16}{81}-4\times\frac{8}{27}=\frac{16}{27}-\frac{32}{27}=-\frac{16}{27}$.

Answer:

$(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(\frac{2}{3},-\frac{16}{27})$