The coordinates of the minimum point of a function can be found using the derivative of the function. For this, we remember that the stationary points have a slope equal to zero. Therefore, we find the roots of the derivative and use the second derivative to confirm whether the point is a minimum.

Here, we will learn about the minimum point of functions. We will learn how to find these points and solve some practice exercises.

## What is the minimum point of a function?

The minimum point of a function is the smallest possible value that we can obtain from the outputs of the function, that is, from the values of *y*. The minimum point is one of the stationary points of a function.

Also, remember that the stationary points of functions are points where the slope of the tangent line is equal to zero. This means that at a minimum point we have $latex \frac{dy}{dx}=0$.

In the following diagram, we can observe the representation of a minimum point of a function:

The slope of the curve is negative on the left-hand side of point P and the slope is positive on the right-hand side. That is, we have the following:

- On the left-hand side of P: $latex \frac{dy}{dx}<0$
- At point P: $latex \frac{dy}{dx}=0$
- On the right-hand side of P: $latex \frac{dy}{dx}>0$

This means that the second derivative of the function at a minimum point is positive since the derivative grows from left to right near that point.

## How to find the minimum point of a function?

We can determine the coordinates of the minimum point of a function by using the derivative of the function to find the stationary points. Then, we use the second derivative to identify which of the points is a minimum.

To find the stationary points, we take the slope of the tangent line at a stationary point to be zero. Then, we form an equation with the derivative and find its roots.

We can follow the following steps to find the minimum point of a function:

**Step** **1:** Find the derivative of the function.

**Step** **2:** Use the derivative of the function to find the stationary points. To do this, we form an equation with the derivative and solve for *x*. That is, we have $latex \frac{dy}{dx}=0$.

**Step** **3:** Determine the nature of the stationary points using the second derivative. When we have a minimum point, we must have $latex \frac{d^2y}{dx^2}>0$.

**Step**** 4:** Use the

*x*-coordinate of the minimum point to find the

*y*-coordinate of the point.

## Minimum point of a function – Examples with answers

**EXAMPLE** **1**

Find the minimum point of the function $latex f(x)=x^2+2x$

##### Solution

**Step 1:** The derivative of the function is:

$latex f(x)=x^2+2x$

$latex f'(x)=2x+2$

**Step** **2:** Using the derivative of the function, we form an equation to find the value of *x*:

$latex 2x+2=0$

$latex 2x=-2$

$latex x=-1$

**Step** **3:** The value of *x* found corresponds to a stationary point. Now, we have to check if the point is a minimum using the second derivative:

$latex f^{\prime \prime}(x)=2$

We got $latex f^{\prime \prime}(x)>0$. This means that the point itself is a minimum point.

**Step**** 4:** We use $latex x=-1$ in the function to find the

*y*-coordinate of the function:

$latex y=x^2+2x$

$latex y=(-1)^2+2(-1)$

$latex y=-1$

The minimum point of the function is (-1, -1).

**EXAMPLE** 2

**EXAMPLE**

Determine the coordinates of the minimum point of the function $latex f(x)=2x^2+8x+2$.

##### Solution

**Step** **1:** We start by finding the derivative of the function. Thus, we have:

$latex f(x)=2x^2+8x+2$

$latex f'(x)=4x+8$

**Step** **2:** We can find the value of *x* by forming an equation with the derivative of the function:

$latex 4x+8=0$

$latex 4x=-8$

$latex x=-2$

**Step** **3:** Using the second derivative to check if the point we find is a minimum point, we have:

$latex f^{\prime \prime}(x)=4$

We have $latex f^{\prime \prime}(x)>0$, so the point itself is a minimum point.

**Step**** 4:** The y-coordinate of the point is:

$latex y=2x^2+8x+2$

$latex y=2(-2)^2+8(-2)+2$

$latex y=-6$

The minimum point of the function is (-2, -6).

**EXAMPLE** 3

**EXAMPLE**

What is the minimum point of the function $latex f(x)=x^3-3x$?

##### Solution

**Step** **1:** We start by finding the derivative of the function:

$latex f(x)=x^3-3x$

$latex f'(x)=3x^2-3$

**Step** **2:** Find the values of *x* by forming an equation with the derivative:

$latex 3x^2-3=0$

$latex x=\sqrt{1}$

$latex x=\pm 1$

$latex x=1~~$ or $latex ~~x=-1$

**Step** **3:** The two values of *x* found correspond to stationary points. So, we use the second derivative to determine what the minimum point is:

$latex f^{\prime \prime}(x)=6x$

When $latex x=1$, we have $latex f^{\prime \prime}(x)>0$ and when $latex x=-1$, we have $latex f^{\prime \prime}(x)< 0$. This means that $latex x=1$ is the minimum point.

**Step**** 4:** The

*y*-coordinate of the minimum point is:

$latex y=x^3-3x$

$latex y=(1)^3-3(1)$

$latex y=-2$

The minimum point of the function is (1, -2).

**EXAMPLE** 4

**EXAMPLE**

Determine the minimum point of the function $latex f(x)=2x^3-6x$.

##### Solution

**Step** **1:** We have the following derivative:

$latex f(x)=2x^3-6x$

$latex f'(x)=6x^2-6$

**Step** **2:** The *x*-values of the stationary points are:

$latex 6x^2-6=0$

$latex x=\sqrt{1}$

$latex x=\pm 1$

$latex x=1~~$ or $latex ~~x=-1$

**Step** **3:** With the second derivative, we determine which of the points is a minimum:

$latex f^{\prime \prime}(x)=12x$

When $latex x=1$, we have $latex f^{\prime \prime}(x)>0$ and when $latex x=-1$, we have $latex f^{\prime \prime}(x)< 0$. This means that $latex x=1$ is the minimum point.

**Step**** 4:** The

*y*-coordinate of the minimum point is:

$latex y=2x^3-6x$

$latex y=2(1)^3-6(1)$

$latex y=-4$

The minimum point of the function is (1, -4).

**EXAMPLE** 5

**EXAMPLE**

Find the minimum point of $latex f(x)=-\frac{1}{3}x^3+x^2+3x$.

##### Solution

**Step** **1:** The following is the derivative of the function:

$latex f(x)=-\frac{1}{3}x^3+x^2+3x$

$latex f'(x)=-x^2+2x+3$

**Step** **2:** We find the *x*-values of the stationary points using the derivative:

$latex -x^2+2x+3=0$

$latex -(x-3)(x+1)=0$

$latex x=3~~$ or $latex ~~x=-1$

**Step 3:** Determine which is the minimum point using the second derivative:

$latex f^{\prime \prime}(x)=-2x+2$

When $latex x=3$, we have $latex f^{\prime \prime}(x)<0$ and when $latex x=-1$, we have $latex f^{\prime \prime}(x)> 0$. This means that $latex x=-1$ is the minimum point.

**Step**** 4:** The

*y*-coordinate of the minimum point is:

$latex y=-\frac{1}{3}x^3+x^2+3x$

$latex y=-\frac{1}{3}(-1)^3+(-1)^2+3(-1)$

$latex y=-\frac{5}{3}$

The minimum point of the function is $latex (-1, ~-\frac{5}{3})$.

**EXAMPLE** 6

**EXAMPLE**

The following right triangle ABC has sides with lengths AB=*x* and BC=*x*+2. Find the minimum area of triangle ABC.

##### Solution

In this exercise, we have to start by finding an equation for the area of the triangle in terms of *x*. Therefore, we use the formula for the area of a triangle with the given lengths:

$latex A=\frac{1}{2}\times BC \times AB$

$latex A=\frac{1}{2}(x+2)x$

$latex A=\frac{1}{2}x^2+x$

Now that we have a function, we can use the steps above.

**Step** **1:** Find the derivative of the function:

$latex A(x)=\frac{1}{2}x^2+x$

$latex A'(x)=x+1$

**Step** **2:** Find the value of *x*:

$latex x+1=0$

$latex x=-1$

**Step** **3:** With the second derivative we verify if the point is a minimum:

$latex A^{\prime \prime}(x)=1$

We have $latex f^{\prime \prime}(x)>0$, so we confirm that the point is a minimum. That is, the area is at its minimum value there.

**Step**** 4:** We use $latex x=-1$ to find the minimum area:

$latex A=\frac{1}{2}x^2+x$

$latex A=\frac{1}{2}(-1)^2-1$

$latex A=-\frac{1}{2}=0.5$

The minimum area of the triangle is 0.5 cm².

## Minimum point of a function – Practice problems

#### What is the minimum point of the function $latex f(x)=2x^3-24x+10$?

Write the coordinates in the input box.

## See also

Interested in learning more about derivatives and stationary points? You can look at these pages:

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