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
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
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
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
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
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:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
