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Econ 407: Further applications of derivatives.

Testing for convexity and concavity using the first derivative. Find the first derivative and solve for critical values by setting the first derivative equal to zero and solving for the root(s). In the second step, find the second derivative and evaluate it for the critical values you found from the first derivative. If the second derivative is:

1. Positive, at the critical point, you have a relative minimum. The function is concave.

2. Negative ,at the critical point, you have a relative maximum. The function is convex. Please examine the following examples and the respective graphs carefully.

Test to see whether the following function is increasing, decresaing, or stationary at x=4.

> y:=x^4-6*x^3+4*x^2-13;

y := x^4-6*x^3+4*x^2-13

> plot(y,x=-2..8);

> dy:=diff(y,x);

> eval(dy,x=4);

Since the first derivative at x=4 is zero, the function is stationary.

Test to see whether the following function is concave or convex at x=3.

> y:=-2*x^3+4*x^2+9*x-15;

y := -2*x^3+4*x^2+9*x-15

> plot(y,x=-3..6);

> dy:=diff(y,x);

dy := -6*x^2+8*x+9

Take the second derivative and evaluate at x=3.

> ddy:=diff(dy,x);

ddy := -12*x+8

> eval(ddy,x=3);

-28

Since the second derivative at x=3 is -28 and negative, the function is concave.

>

> plot({y,-6},x=-3..4);

> eval(y,x=3);

-6

Find the critical values of the following function and test to see if at the critical points the function is a relative maximum, minimum, or possible inflection point.

> y:=-(x-8)^4;

y := -(x-8)^4

> dy:=diff(y,x);

dy := -4*(x-8)^3

> solve(dy,x);

8   8   8

The critical points are 8, 8, 8.

> ddy:=diff(dy,x);

ddy := -12*(x-8)^2

> eval(ddy,x=8);

>

Since the second derivative is zero, the test is inconclusive. The fourth derivative of the function evlauted at x=8 is negative, the function is concave and at a relative maximum. See the graph below:

> plot(-(x-8)^4, x=6..10);

>

Optimize the following functions by finding the critical values. Make sure you use the second derivative test to see if you have a relative maximum or minimum.

> y:=7*x^2+112*x-54;

>