Single Variable Calculus

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Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 66

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Determine the intervals of increase or decrease, the intervals of concavity and the points of inflection of $\displaystyle f(x) = \frac{e^x}{x^2}$.

$ \begin{equation} \begin{aligned} \text{if } f(x) =& \frac{e^x}{x^2}, \text{then by using Quotient Rule..} \\ \\ f'(x) =& \frac{x^2 (e^x) - e^x (2x)}{(x^2)^2} = \frac{xe^x (x - 2)}{x^4} = \frac{e^x (x - 2)}{x^3} \end{aligned} \end{equation} $

Again, by using Quotient Rule as well as Product Rule..

$ \begin{equation} \begin{aligned} f''(x) =& \frac{x^3 [e^x (1) + e^x (x - 2)] - [e^x(x - 2)] (3x^2) }{(x^3)^2} \\ \\ f''(x) =& \frac{x^2 [x^2 e^x - xe^x - 3xe^x + 6e^x]}{x^6} \\ \\ f''(x) =& \frac{x^2 e^x - 4xe^x + 6ex}{x^4} \\ \\ f''(x) =& \frac{e^x (x^2 - 4x + 6)}{x^4} \end{aligned} \end{equation} $

Now, to determine the intervals of increase or decrease, we must get first the critical numbers by setting $f'(x) = 0$. So,

$\displaystyle f'(x) = \frac{e^x (x - 2)}{x^3}$

when $f'(x) = 0$

$\displaystyle 0 = \frac{e^x (x - 2)}{x^3}$

The real solution is..

$x = 2$

Hence, the interval of increase or decrease is..

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f \\ \hline\\ x < 2 & - & \text{decreasing on } (- \infty, 2 ) \\ \hline\\ x > 2 & + & \text{increasing on } (2, \infty)\\ \hline \end{array} $

Next to determine the inflection points, we set $f''(x) = 0$. So,

$\displaystyle 0 = \frac{e^x (x^2 - 4x + 6)}{x^4}$

It shows that we have no inflection point because we don't have real solution for the equation. Let's evaluate $f''(x)$ with interval..

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity} \\ \hline\\ x < 0 & - & \text{Downward} \\ \hline\\ x > 0 & - & \text{Downward}\\ \hline \end{array} $

The function has downward concavity at $(- \infty, 0)$ and $(0, \infty)$.

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