graph the function y = 2x^4 + 4x^3 = 2x^3(x + 2) by identifying the domain and any symmetries, finding the…

Explanation:

Step1: Find the first - derivative

Given $y = 2x^{4}+4x^{3}=2x^{3}(x + 2)$. Using the product - rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = 2x^{3}$ and $v=x + 2$, or simply expanding $y=2x^{4}+4x^{3}$ and differentiating term - by - term. The derivative $y^\prime=8x^{3}+12x^{2}=4x^{2}(2x + 3)$.

Step2: Find the critical points

Set $y^\prime = 0$. So $4x^{2}(2x + 3)=0$. Solving this equation gives $x = 0$ and $x=-\frac{3}{2}$.

Step3: Analyze the sign of the first - derivative

Choose test points in the intervals $(-\infty,-\frac{3}{2})$, $(-\frac{3}{2},0)$ and $(0,\infty)$. For $x=-2$ (in the interval $(-\infty,-\frac{3}{2})$), $y^\prime=4(-2)^{2}(2(-2)+3)=16(-1)=-16<0$. For $x =-\frac{1}{2}$ (in the interval $(-\frac{3}{2},0)$), $y^\prime=4(-\frac{1}{2})^{2}(2(-\frac{1}{2})+3)=4\times\frac{1}{4}(2)=2>0$. For $x = 1$ (in the interval $(0,\infty)$), $y^\prime=4(1)^{2}(2(1)+3)=4\times5 = 20>0$. Since the function changes from decreasing ($y^\prime<0$) to increasing ($y^\prime>0$) at $x =-\frac{3}{2}$, there is a local minimum at $x =-\frac{3}{2}$. Substitute $x =-\frac{3}{2}$ into the original function $y = 2(-\frac{3}{2})^{4}+4(-\frac{3}{2})^{3}=2\times\frac{81}{16}+4\times(-\frac{27}{8})=\frac{81}{8}-\frac{27}{2}=\frac{81 - 108}{8}=-\frac{27}{8}=- 3.4$ (rounded to the nearest tenth). The function does not change sign at $x = 0$ (since $y^\prime$ has a double - root at $x = 0$), so there is no local maximum or minimum at $x = 0$.

Step4: Find the second - derivative

Differentiate $y^\prime=8x^{3}+12x^{2}$ to get $y^{\prime\prime}=24x^{2}+24x=24x(x + 1)$.

Step5: Analyze concavity and inflection points

Set $y^{\prime\prime}=0$, then $24x(x + 1)=0$, which gives $x = 0$ and $x=-1$ as inflection points.

Answer:

A. The local minimum/minima is/are located at $(-1.5,-3.4)$ B. There are no local maxima.