Graph the function. Label all intercepts, critical points, and inflection points.

Answer

• Find dots.

The function *f* is undefined when *x* = 1, and has a vertical asymptote there.

Before finding derivatives, rewrite the original function:

*f* (*x*) = 4(*ln* *x*)^{-1}

The derivative is

This is never undefined or 0 at a place where *f* is defined, so the function *f* has no critical points.

The second derivative is

This is undefined at the same places as *f* and *f*'. This is zero when

The corresponding *y* value of the inflection point is

For the sake of graphing,

*e*^{-2} ≅ 0.135.

Here are the points:

• Find shapes.

We'll start with the numberlines.

The function *f* is negative when *x* < 1 and positive when *x* > 1.

The derivative

*f*'(*x*) = -4*x*^{-1}(*ln* *x*)^{-2}

is always negative, so *f* is always decreasing.

The second derivative is

The factors 4*x*^{-2} and (*ln* *x*)^{-2} are always positive, so the sign of *f*" is determined by the sign of the factor [1 + 2(*ln* *x*)^{-1}].

When *x* > 1, the quantity *ln* *x* is positive and so is [1 + 2(*ln* *x*)^{-1}].

When *x* < 1, we know there's a possible inflection point at *e*^{-2} ≅ 0.135. We need to know what's going on with the sign of *f*" on the intervals

0 < x < *e*^{-2}

and

*e*^{-2} < *x* < 1.

In order to not find too much of a headache, plug in some numbers and find the sign of [1 + 2(*ln* *x*)^{-1}]. Since *e*^{-2} ≅ .135, we'll use *x* = 0.1 for the first interval and *x* = 0.5 for the second. When *x* = 0.1, we find

[1 + 2(*ln* .1)^{-1}] ≅ 0.131 > 0

so *f*" is greater than zero on the interval 0e^{{-2}}. When *x* = .5 we find[1 + 2(*ln* .5)^{-1}] ≅ -1.885 < 0

so *f*" is less than zero on the interval *e*^{{-2}}

This means *f* is concave up at the beginning of the graph, then concave down until the asymptote, then concave up again. Here are the shapes:

• Play connect-the-dots.

Here's the final graph: