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Sketching Curves

April 10th, 2009 |

Author: Merrin

Sketching curves is an important application of differentiation. The easier graphs contain less features such as discontinuities, asymptotes, or points of inflection. We can begin learning how to sketch curves by constructing a simple example dealing with parabolas that we already know how to plot and move onto a more advanced graph containing more features.

Example 1. Consider two parabolas $latex f(x) = x^2$ and $latex g(x) = -(x-a)^2 + c$. One parabola is open upwards and the other parabola is open downwards shifted to the right by $latex a$ and shifted up by $latex c$. Solve for $latex a$ and $latex c$ such that a new function is continuous and the derivative is continuous.

$\begin{array}{l} h(x) = f(x) \quad x\le2 \\ h(x) = g(x) \quad x \ge 2 \end{array}$

Solution 1. Since the function is continuous $latex x^2 = -(x-a)^2 + c$ and since the derivative is continuous $latex 2x = -2(x-a)$. From this equation we know that the crossover has to occur at $latex x=a/2$. Since the function is also continuous we have

$\cfrac{a^2}{4} = -\cfrac{a^2}{4} + c$

$c = \cfrac{a^2}{2}$

Everything is well defined for a given choice of $latex a=2$, the shift of the second parabola to the right, which then determines $latex c=2$. doubleparabola

Two parabolas can be connected together to form a curve which is continuous and has a continuous derivative

There are several features of this graph that if you understand can really help you plot. Convince yourself that the slope is monotonically increasing up to x=1. After this point convince yourself that the slope is monotonically decreasing. The point x=1 is a special point where there is a crossover in how the slope changes called a point of inflection. When the slope is always increasing that part of the curve is called concave up, when the slope is decreasing that part of the curve is called concave down. A full concave up curve can hold some water like a cup. A concave up curve that has a slope that changes signs from negative to zero to positive has a local minimum where the slope is zero. Local maximum occur on a concave down curve that switches slope form positive to zero to negative. At the transition between concave up and concave down their is what is called a point of inflection to identify the change. Concave up occurs when the second derivative is positive. Concave down occurs when the second derivative is negative. Inflection can occur where the second derivative is zero. If you plot all three $latex f(x)$, $latex f’(x)$, and $latex f”(x)$ then you can also see how all the analysis plays out.

dvaparabola1

Information can be gathered by looking at the first and second derivative in conjunction with the original graph

Critical points are where singularities occur, the endpoints of the domain, or locations where the slope is zero. Evaluating the function at all the critical points can be used to determine the global maximum and the global minimum. This is very useful for optimization problems. As we have seen this all derives from knowing how to graph functions using our calculus skills.

Example 2. Sketch the function $latex f(x) = x^2e^{-x}$ over the interval $latex x \in (-\infty, \infty) $. Indicate the locations of any critical points or points of inflection. Specify where the function is concave up or concave down.

Solution 2. There are a lot of details we have to analyze so we might as well get the derivatives out of the way first.

$f(x) = x^2e^{-x}$

$f'(x) = (2x-x^2)e^{-x}$

$f''(x) = (2-4x+x^2)e^{-x}$

Since $latex e^{-x}$ is always positive the prefactor polynomials determine all the action. The critical points from the first derivative are where $latex 2x=x^2$, or x=0 and x=2. We can search for the points of inflection be solving

$x^2-4x+2=0$

$x = 2 \pm \sqrt{2}$

These points tell us where the curve shifts from concave up to concave down. The curve is concave up to the left of $latex 2+sqrt{2}$ is concave down between $latex 2-\sqrt{2}$ and $latex 2+\sqrt{2}$ and then concave up to the left of that interval. Next we solve for any asymptotes. There is one at $latex x=\infty$ the asymptote is $latex y =0$ for negative infinity the function increases unbounded. Some key points are to evaluate the function at the critical points, points of inflection, and y intercept where f(x)=0. As you can see many details of the graph can be revealed by simple calculus calculations. Here is the graph

exgraph2