Question

Describe what the first-derivative test is for and how to use it. Sketch graphs and sign charts to illustrate your description.

Short answer

The first derivative test states that $f$is continuous and differentiable at every point on$(a,b)$.

Step-by-step solution

Step 1. Given Information.

Using the first derivative test and showing the graphs for the various descriptions.

Step 2. First derivative test.

Suppose $x=c$is the location of a critical point of a function $f$, and let $(a,b)$be an open interval around $c$that is contained in the domain of $f$and does not contain any other critical points of $f$.

If $f$is continuous on $(a,b)$and differentiate at every point of $(a,b)$except possibly at $x=c$, then the following statements hold.

Step 3. Graphical Representation for the various ways.

(a) If $f\text{'}\left(x\right)$is positive for $x?(a,c)$and negative for $x?(c,b)$, then $f$has a local maximum at $x=c$.

The first derivative $f\text{'}$changes from positive to negative at $x=c$.

(b) If $f\text{'}\left(x\right)$is negative for $x?(a,c)$and positive for $x?(c,b)$, then $f$has a local minimum at $x=c$.

The first derivative $f\text{'}$changes from negative to positive at $x=c$.

(c) If $f\text{'}\left(x\right)$is positive for both $x?(a,c)$and $x?(c,b)$, then $f$does not have a local extremum at $x=c$.

The first derivative $f\text{'}$is positive on both sides of $x=c$.

(d) If $f\text{'}\left(x\right)$is negative for both $x?(a,c)$and $x?(c,b)$, then $f$does not have a local extremum at $x=c$.

The first derivative$f\text{'}$is negative on both sides of$x=c$.