Question

Sketch the graph of the function. Choose a scale that allows all relative extrema and points of inflection to be identified on the graph. \(y=3 x^{4}+4 x^{3}\)

Short answer

The graph peaks at x=0 and x=-1, with points of inflection at x=0 and x=-2/3

Step-by-step solution

Find the First Derivative

First find the derivative (\(y'\)) of the function \(y=3x^{4}+4x^{3}\) using the power rule for differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\). The first derivative of the function is: \(y'=12x^{3}+12x^{2}\)

Find the Critical Points

The critical points occur where the first derivative (\(y'\)) is equal to zero or undefined. For this function, the derivative is never undefined. So, set the derivative equal to zero and solve for \(x\): \(0=12x^{3}+12x^{2}\) --> \(0=x^{2}(12x+12)\). Therefore the critical points are \(x=0\) and \(x=-1\)

Find the Second Derivative

Differentiate the equation for the first derivative once more using the power rule to get the second derivative (\(y''\)): \(y''=36x^{2}+24x\)

Find the Points of Inflection

Set the second derivative equal to zero and solve for \(x\) to find the points of inflection: \(0=36x^{2}+24x\) --> \(0=12x(3x+2)\). Therefore the points of inflection are \(x=0\) and \(x=-2/3\).

Sketch the Graph

Use the critical points and points of inflection to sketch the graph. Label these points on the graph and pay attention to how the behavior of the function changes at these points. For \(x>0\), the function is decreasing and concave up. For \(-1<x<0\) the function is increasing and concave up, while for \(x<-1\) it's increasing and concave down. Recognize the symmetry of derivative sign changes around the points of inflection.