Q. 85 - Calculus

Question

In Exercises 83–86, use the given derivative $f\text{'}$ to find any local extrema and inflection points of f and sketch a possible graph without first finding an formula for f .

$f\text{'}\left(x\right) = \frac{1}{x}.$

Step-by-Step Solution

Answer

The function does not have any local extremum and below given is the graph of it.

Step 1. Given Information.

Given is the first derivative of the function:

$f\text{'}\left(x\right) = \frac{1}{x}.$

Step 2. Theorem.

The Derivative Measures Where a Function is Increasing or Decreasing

Let f be a function that is differentiable on an interval I.

(a) If $f\text{'}$ is positive in the interior of I, then f is increasing on I.

(b) If $f\text{'}$ is negative in the interior of I, then f is decreasing on I.

(c) If $f\text{'}$ is zero in the interior of I, then f is constant on I.

Step 3. First derivative test.

To find the local extrema the first derivative of the function must be zero. So,

$$\begin{gathered} f\text{'}\left(x\right) = 0, \\ \frac{1}{x} = 0, \end{gathered}$$

As we see x will be not defined it $f\text{'}\left(x\right)$ is equal to zero.

This means the function does not have any local extrema.

Step 4. Second derivative test.

$$\begin{gathered} Differentiating f\text{'}\left(x\right) we get, \\ f\text{'}\text{'}\left(x\right) = -\frac{1}{x^2}, \end{gathered}$$

Inflection point occurs when $f\text{'}\text{'}\left(x\right) = 0,$ but as we can see that $f\text{'}\text{'}\left(x\right)$ cannot be zero as if it becomes zero then x will not be defined.

So, this shows us that there is no inflection point where concavity changes.

And also from the above equation we can see that $f\text{'}\text{'}\left(x\right)$ is always negative as $x^2$ cannot be negative. This means the concavity of f will be downwards.

Step 5. Drawing the graph.

Keeping in mind the above steps let's draw the graph of the function.