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

graph the function y = 2x^4 + 4x^3 = 2x^3(x + 2) by identifying the domain and any symmetries, finding the derivatives y and y, finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. find the derivative y. y = find the second derivative y. y = identify any critical points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

graph the function y = 2x^4 + 4x^3 = 2x^3(x + 2) by identifying the domain and any symmetries, finding the derivatives y and y, finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. find the derivative y. y = find the second derivative y. y = identify any critical points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Answer

Explanation:

Step1: Find the first - derivative

Use the power rule $(x^n)'=nx^{n - 1}$ and the sum - rule $(u + v)'=u'+v'$. For $y = 2x^{4}+4x^{3}$, we have $y'=(2x^{4})'+(4x^{3})'=8x^{3}+12x^{2}=4x^{2}(2x + 3)$.

Step2: Find the second - derivative

Differentiate $y'=8x^{3}+12x^{2}$ with the power rule. $y''=(8x^{3})'+(12x^{2})'=24x^{2}+24x=24x(x + 1)$.

Step3: Find critical points

Set $y' = 0$. $4x^{2}(2x + 3)=0$. Using the zero - product property, $4x^{2}=0$ gives $x = 0$ and $2x+3=0$ gives $x=-\frac{3}{2}$.

Answer:

$y'=8x^{3}+12x^{2}$ $y''=24x^{2}+24x$ Critical points: $x = 0,x=-\frac{3}{2}$