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Show that for any real numbers, A, B the function f(x) = \ln ((x + A)(x + B)) never changes...

Question:

Show that for any real numbers, A, B the function {eq}f(x) = \ln ((x + A)(x + B)) {/eq} never changes concavity.

Concavity of a Function:

We consider a function of one real variable {eq}f(x). {/eq}

The concavity of the function can be studied by analyzing the sign of the second derivative.

The function is concave up in the intervals where the second derivative is positive and concave down in the

intervals where the second derivative is negative.

Answer and Explanation:

Given the function

{eq}\displaystyle f(x) = \ln ((x + A)(x + B)) = \ln(x+A) + \ln(x+B) {/eq}

where A and B are some real constants, the second derivative of the function is found as

{eq}\displaystyle f'(x) = \frac{1}{x+A} + \frac{1}{x+B} \\ \displaystyle f''(x) = -\frac{1}{(x+A)^2} - \frac{1}{(x+B)^{2}} \\ \displaystyle = -\frac{(x+B)^{2}+(x+A)^{2}}{(x+A)^2(x+B)^{2}}. {/eq}

As can be observed, the second derivative is always a negative quantity for any value of the constants A and B.

Consequently, the function never changes the concavity.


Learn more about this topic:

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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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