Question

Sketch the graph of a function that is continuous on $(-\mathrm{\infty },\mathrm{\infty })$ and satisfies the following sets of conditions. $$\begin{array}{rl}& f^{\prime \prime }(x)>0\text{ on }(-\mathrm{\infty },-2);f^{\prime \prime }(-2)=0;f^{\prime }(-1)=f^{\prime }(1)=0\\ & f^{\prime \prime }(2)=0;f^{\prime }(3)=0;f^{\prime \prime }(x)>0\text{ on }(4,\mathrm{\infty })\end{array}$$

Short answer

Short Answer: The function's graph is concave up on the intervals $(-\mathrm{\infty },-2)$ and $(4,\mathrm{\infty })$, with inflection points at $x=-2$ and $x=2$. It has three critical points at $x=-1$, $x=1$, and $x=3$, which are assumed to be local maximum, local minimum, and local maximum, respectively. By connecting these points with smooth curves, the graph maintains appropriate concavity for each interval.

Step-by-step solution

Interpret the given conditions

Let's analyze the conditions given for $f^{\prime }(x)$ and $f^{″}(x)$: - $f^{\prime \prime }(x)>0$ on $(-\mathrm{\infty },-2)$: This means the function is concave up on this interval because the second derivative is positive. - $f^{\prime \prime }(-2)=0$: This means that there is an inflection point at $x=-2$ because the second derivative changes from positive to either negative or zero. - $f^{\prime }(-1)=f^{\prime }(1)=0$: These are the critical points of the function, which indicate either a local minimum, local maximum, or neither. - $f^{\prime \prime }(2)=0$: This means that there is another inflection point at $x=2$ because the second derivative changes from negative or zero to positive. - $f^{\prime }(3)=0$: This is another critical point. - $f^{\prime \prime }(x)>0$ on $(4,\mathrm{\infty })$: This means the function is concave up on this interval because the second derivative is positive.

Combine the conditions for the overall shape

Now that we have analyzed the conditions, we can outline the overall behavior of the function based on those conditions. - The function is concave up on $(-\mathrm{\infty },-2)$ and $(4,\mathrm{\infty })$. - The function has inflection points at $x=-2$ and $x=2$. - The function has critical points at $x=-1$, $x=1$, and $x=3$.

Sketch the graph

To sketch the graph, we will begin by drawing the inflection points and critical points on the coordinate plane: - Plot the points $(-2,0)$, $(2,0)$, $(-1,0)$, $(1,0)$, and $(3,0)$. Next, remember that the function is concave up on $(-\mathrm{\infty },-2)$ and $(4,\mathrm{\infty })$: - For $x<-2$, draw a smooth curve that is concave up, connecting the point $(-2,0)$. - For $x>4$, draw a smooth curve that is also concave up, starting from the point $(3,0)$. Now, connect the inflection points and critical points with smooth curves that satisfy the concave up or down behavior: - For \(-24\). Your sketch should now display a continuous function that satisfies the given conditions. An example can be looked up and shown for comparison, but note that the exact shape of the graph may vary slightly, as long as the given conditions are met.

Key concepts

Concave Up Function

When we talk about a concave up function, we're describing the shape of the graph where it curves upwards like the open side of a cup. Imagine bending a flexible rod so that the ends rise higher while the middle dips lower—that's the concave up shape.

To be technical, a graph is concave up on an interval where its second derivative, denoted as $f^{″}(x)$, is positive. This means that the slope of the function is increasing as you move from left to right across the interval. In the context of the exercise, the given conditions suggest that the function is concave up for $x<-2$ and $x>4$, thus the corresponding parts of the graph should appear like a hill that's being elevated.

Inflection Points

An inflection point occurs where a curve changes its concavity, that is, from concave up to concave down, or vice versa. Think of it as the point where a hill turns into a valley or the other way around. Mathematically, it's typically found where the second derivative of the function is equal to zero, denoted as $f^{″}(x)=0$, and where there's an actual change in concavity.

The exercise specifies that there are inflection points at $x=-2$ and $x=2$. These points signify where the curvature of our graph switches, and this change in behavior is crucial to sketching an accurate graph of the function.

Critical Points

Critical points are the x-values where the function's first derivative, $f^{\prime }(x)$, is either zero or undefined. They are important for determining the location of potential local maxima and minima, essentially the peaks and troughs, of the function.

In our example, $f^{\prime }(x)$ is zero at $x=-1$, $x=1$, and $x=3$. Knowing this, we mark these values as critical points. Critical points help us sketch the general outline of a graph by indicating where the function might change direction. By incorporating this information, students can visualize the function's behavior and place breakpoints on their graph with greater accuracy.

Second Derivative Test

The Second Derivative Test is a handy tool that helps us determine whether a critical point is a local maximum, local minimum, or neither—a saddle point. Simply put, if the second derivative at a critical point is positive, $f^{″}(x)>0$, the function has a local minimum there. Conversely, if the second derivative is negative, $f^{″}(x)<0$, there's a local maximum. If it's zero, the test is inconclusive, and we may need to use other methods to find out more about the point.

For the given exercise, the second derivative test isn't directly applicable to every critical point because the second derivative is zero at some points. However, we use the test indirectly by interpreting the given conditions—since the function transitions from concave up to concave down or from concave down to concave up at our critical points, we get an indication of whether these points are maxima or minima.