Analyze and sketch the graph of the function

Question

Analyze and sketch the graph of the function. Identify any relative extrema, points of inflection, and asymptotes. Round answers to three decimal places. If answer does not exist enter DNE.

y= ln(7x)/x²

relative minimum (x,y) =

relative maximum (x,y) =

inflection point (x,y) =

vertical asymptote x =

horizontal asymptote y=

Graph the function

Answer 1

First, to figure out the behavior of the graph, always a good starting point.

As x approaches 0 from the positive side, y goes to -∞. An easy way to see this is to use l'Hopital's rule.

The derivatives of the numerator and denominator give you 1/x/(-1/x³) = -1/x², which goes to -∞. At x = 1/7, 7x = 1 and ln(7x) = 0, so this is the y-intercept. The function will continue increasing but then decrease to 0, as again using l'Hopital's rule as x -> ∞, -1/x², -> 0.

From the behavior as described above, we see that there is no relative minimum.

For the relative maximum, take the first derivative and set it equal to 0

f'(x) = (1/x(x²) - 2x ln(7x))/x⁴ =

(x- 2x ln(7x))/x⁴ =

(1 - 2 ln(7x))/x³

Setting the derivative to 0 by settiing the numerator to 0,

1 = 2ln(7x)

1/2 = ln(7x)

e^1/2 = 7x

x = e^1/2/7

ln(7x)/x² = 1/2/(e/49) = 49/2e

(x,y) = (e^1/2/7,49/2e)

I would graph the function from .12 to 10

f''(x) = (6ln(7x) - 5)/x⁴

This equals 0 when the numerator equals 0

6ln(7x) - 5 = 0

6ln(7x) = 5

ln(7x) = 5/6

7x = e^5/6

x = e^5/6/7

y = ln(7x)/x² = 5/6/(e^5/3/49) = 245/(6e^5/3)

The vertical asymptote is x = 0.

The horizontal asymptote is y= 0