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. $y=x^4-2x^2$

Short answer

The function $y=x^4-2x^2$ has a local maximum at x=0 and local minima at x=1 and x=-1.

Step-by-step solution

Find the derivative

First, differentiate the function $y=x^4-2x^2$ to get $y^{\prime }=4x^3-4x$. The derivative will be used to find critical points (where the function increases or decreases) and points of inflection (where concavity changes).

Find the critical points

To find the critical points, set the derivative equal to zero and solve for $x$. So, $4x^3-4x=0$. Factor out the common factor $4x$, we get $x(1-x^2)=0$. Setting this equal to zero gives $x=0$, $x=1$, and $x=-1$. These are our critical points.

Determine the nature of the critical points

To determine whether these points are local minima, maxima, or points of inflection, we'll perform the second derivative test. Compute the second derivative of $y$, denoted as $y^{″}$, which equals $12x^2-4$. Plug the critical points into $y^{″}$. If $y^{″}>0$, then the point is a local minimum. If $y^{″}<0$, then the point is a local maximum. For $x=0$, $y^{″}=-4<0$, so $x=0$ is a local maximum. For $x=1$ and $x=-1$, $y^{″}=8>0$, so both are local minimums.

Sketch the graph

Having determined the critical points and their nature, we can now sketch the graph of $y=x^4-2x^2$. Mark the points (0,0), (1,-1), and (-1,-1) on the graph. Draw the graph in such a way that it has a local maximum at x=0 and local minima at x=1 and x=-1, and passes through the origin.