Analyze the given function using calculus f(x)=(1/x)+ln(x)

Question

a) Using the first derivative only, identify the intervals where the function increases or decreases.

b) Using the second derivative only, identify the intervals where the function concaves upward or downward.

c) Identify all inflection points of the function.

Answer 1

Some people would say Step 1 is to take the derivative. I would say that a better Step 1 is to consider the domain of the function. In doing so, you see that ln(x) only exists for x > 0

Step 2. Take the derivative (1st derivative)

It might be helpful to think of 1/x as x^-1 then to use the power rule meaning you move the exponent out front and reduce the exponent by 1. That would give you -x^-2 or -1/x². Then the derivative of ln(x) is 1/x (the logarithm rule). That makes f '(x) = 1/x - 1/x²

Step 3. Set the derivative equal to zero and solve

(also find values of "x" where the derivative does not exist)

1/x - 1/x² = 0

1/x(1 - 1/x) = 0

so either 1/x = 0 or 1 - 1/x = 0

There are no values of x that make 1/x = 0 but x = 0 is a place the derivative does not exist.

1 - 1/x = 0

1 = 1/x

x = 1

So x = 0 (the place where the derivative DNE) and x = 1 are the critical points.

Step 4. Put the critical points on a number line.

Then calculate the value of the 1st derivative in each section of the number line by picking an easy value of x in each interval and plugging it into the 1st derivative function. Note: you only need to determine if the value of the first derivative function is positive or negative. You don't necessarily need to know the exact value.

In this case, since the domain of the function is x > 0, we don't need to consider the interval left of zero.

In the region between 0 and 1, we can pick x = 1/2. Plugging in 1/2 as "x" in f ' gives us:

1/(1/2) - 1/(1/2)² = 2 - 4 = -2 so we put a negative symbol on this interval.

In the region greater than 1, we can pick x = 2. Plugging in 2 as "x" in f ' gives us:

1/2 - 1/2² = 1/2 - 1/4 = 1/4 so we put a positive symbol on this interval.

Looking at the number line, the regions with a negative are regions where the function is decreasing. The regions with a positive are regions where the function is increasing.

Step 5. Take the second derivative:

Since f '(x) = x^-1 - x^-2

f ''(x) = -x^-2 + 2x^-3 or f ''(x) = 2/x³ - 1/x²

Step 6. Set the 2nd derivative equal to zero and solve.

(also find values of "x" where the 2nd derivative does not exist)

2/x³ - 1/x² = 0

1/x²(2/x - 1) = 0

So either 1/x² = 0 or 2/x - 1 = 0

There are no values of x that make 1/x² = 0 but x = 0 is a place the 2nd derivative does not exist.

For 2/x - 1 = 0, x = 2

Step 7. Place the solutions above on a number line

Calculate the value of the 2nd derivative in each section

A number in the region between 0 and 2 is x = 1. Plugging in x = 1 into f '' gives us:

2/1³ - 1/1² = 2 - 1 = 1 (a positive)

A number in the region greater than 2 is x = 3. Plugging in x = 3 into f '' gives us:

2/3³ - 1/3² = 2/27 - 1/9 = -1/27 (a negative)

Regions with a positive 2nd derivative value are concave up. Regions with a negative are concave down. Use open intervals because the function is neither concave up or concave down at the values of 0 and 2.

If the concavity changes, the boundary point is a point of inflection.

Answer 2

f(x) = 1/x + lnx

f^'(x) = - 1/x² + 1/x = 0

-1/x ( 1/x - 1 ) = 0

x = 1 f^'(x) < 0 for 0 < x < 1 and f^'(x) > 0 for x > 1

f decreasing on (0 , 1) and increasing on (1 , ∞ ).

f^"(x) = 2/x³ - 1/x² = 0

1/x² (2/x - 1) = 0

x = 2 f^"(x) > 0 for 0 < x < 2 and f^"(x) < 0 for x > 2

f concave up on (0 , 2) and concave down on (2 , ∞ ).