Question

In Exercises 62 and \(63,\) feel free to use a CAS (computer algebra system), if you have one, to solve the problem. Logistic Functions Let \(f(x)=c /\left(1+a e^{-h x}\right)\) with \(a>0\) \(a b c \neq 0\) (a) Show that \(f\) is increasing on the interval \((-\infty, \infty)\) if \(a b c>0\) and decreasing if \(a b c<0\) . (b) Show that the point of inflection of \(f\) occurs at \(x=(\ln |a|) / b\)

Short answer

Based on the analysis and solution procedures, the function \(f(x) = c / (1 + a e^{-hx})\) is increasing over the domain if \(abc > 0\) and decreasing if \(abc < 0\). Also, the point of inflection for the given function is at \(x = (\ln |a|) / b\).

Step-by-step solution

Find the derivative of \(f\)

First, it's crucial to find the derivative of \(f(x)\), termed as \(f'(x)\), which measures the rate at which \(f(x)\) changes for every unit increase in \(x\). This can be achieved using the quotient and chain rule for differentiation. Remember to simplify the result.

Discuss the sign of the derivative

Next, we will discuss the sign of the derivative under the conditions \(abc > 0\) and \(abc < 0\). If \(abc > 0\), then by multiplying the factors of the derivative, we know that \(f'(x)\) is always positive, hence \(f(x)\) is increasing over the entire domain. On the other hand, if \(abc < 0\), \(f'(x)\) will always be negative, indicating that \(f(x)\) is decreasing everywhere.

Find the second derivative of \(f\)

Now we need to find the second derivative of \(f(x)\), termed as \(f''(x)\), which reflects the rate and direction of change in \(f'(x)\). This can be found by differentiating \(f'(x)\) using the quotient and chain rule again.

Determine the point of inflection

We will have a point of inflection when the function changes concavity, which is exactly when \(f''(x) = 0\). We thus solve the equation \(f''(x) = 0\) for \(x\), and after some algebraic simplification, it should be found that \(x = (\ln|a|)/b\).