Subject: Calculus

# First Derivative Test

Possible locations of critical points in a function's graph.

In the previous article on stationary point, we learned how to find the stationary and the critical points in the graph of a function. We also know that a critical or stationary point is either a point that sits at the "peak" of the curve of that function (commonly known as a local maximum), a point that rests on the "valley" of the curve of that function (also known as local minimum) and a point of discontinuity in the function. However, we don't know exactly what point sits on the "valley", rests on the "peak" or is a point of discontinuity. Thus, it is our goal to seek answer to these questions.

## The first derivative test

The process of determining whether a given critical point (or stationary point) of a function is a local maximum, a local minimum, or neither is called first derivative test. As its name suggests, it differentiates the given function once and see whether a certain critical point is a local minimum, a local maximum or neither. Simply, if the value of the first derivative is negative at the points to the left of the critical point, and positive to the right of it, it is a relative minimum. If value of the first derivative is positive at the points to the left of the critical point, and negative to the right of it, the critical point is a relative maximum. Let's take this one step-by-step.

### First Derivative Test(Guidelines in solving for the local maximum or minimum)

• Solve for the first derivative of the function f. That is, find f'(x).
• Solve for the critical values(x_{cn}) of f(x). That can be done by solving x when f'(x)=0 or f'(x) does not exist. Label the smallest critical value as x_{c1} and the largest critical value as x_{cn}, n=1,2,3,.. and so on.
• For each critical value(x_{cn}, choose a point x_{l} in the function f(x) that lies to the left of the critical value such that x_{l}>x_{c(n-1)}.
• For each critical value(x_{cn}, choose a point x_{r} in the function f(x) that lies to the right of the critical value such that x_{r}<x_{c(n+1)}.
• Evaluate f(x=x_{l}) and f(x=x_{r}).
• If f(x=x_{l}) <0 and f(x=x_{r})>0, then the critical point corresponds to the local minimum. Likewise, if f(x=x_{l}) >0 and f(x=x_{r})<0, then that critical point corresponds to the local maximum.

Remark: If f(x=x_{l})>0 and f(x=x_{r})>0, or f(x=x_{l})<0 and f(x=x_{r})<0, then the function has no local maximum or minimum relative to that interval.

### Example #1

Find the local maximum and local minimum of the following function.

f(x)=3x^{2}-\frac{3}{4} x^{4}

Differentiating the above equation gives

• f(x)=6x-4\cdot \frac{3}{4}x^{3}

f'(x)=6x- 3x^{3}

• Solving for the critical points, we have

f'(x)=0
6x-3x^{3}=0
critical values:x=0,\sqrt{2},-\sqrt{2}

• at the critical point 0,
to the left of 0: f'(x)<0
to the right of 0: f'(x)>0
• at the critical point -\sqrt{2},
to the left of -\sqrt{2}: f'(x)>0
to the right of -\sqrt{2}: f'(x)<0
• at the critical point \sqrt{2},
to the left of \sqrt{2}: f'(x)>0
to the right of \sqrt{2}: f'(x)<0

Thus, the critical point 0 corresponds to the local minimum point and the points -\sqrt{2} and \sqrt{2} corresponds to the local maximum point.

NEXT TOPIC: Second Derivative Test