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Magazine Chapter 15: Problem 341
Locate the maxima and minima of \(\mathrm{y}=2 \mathrm{x}^{2}-8 \mathrm{x}+6\)
Short Answer
Expert verified
The function \(y = 2x^2 - 8x + 6\) has a local minimum at the point (2, -2). There are no local maxima for this function since it is a parabola opening upwards.
Step by step solution
01
Differentiate the quadratic function with respect to x
To find the critical points of the function \(y = 2x^2 - 8x + 6\), we first need to find its derivative with respect to x. Using the power rule, we have:
\[
\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 8x + 6) = 4x - 8
\]
02
Set the first derivative equal to zero
Now that we have the first derivative, we need to find the critical points by setting the derivative equal to zero:
\[
4x - 8 = 0
\]
03
Solve for x
Solving the equation from the previous step, we get:
\[
x = \frac{8}{4} = 2
\]
04
Analyzing the second derivative
To determine whether x=2 corresponds to a local maximum or minimum, we'll examine the sign of the second derivative. First, let's compute the second derivative:
\[
\frac{d^2y}{dx^2} = \frac{d}{dx}(4x - 8) = 4
\]
Since the second derivative is positive at all points (including x=2), the function is concave up, which means that x=2 corresponds to a local minimum.
05
Calculate the value of the minimum
Now that we know that x=2 corresponds to a local minimum, let's find the value of the function at that point:
\[
y = 2(2)^2 - 8(2) + 6 = 8 - 16 + 6 = -2
\]
So the local minimum occurs at the point (2, -2). There are no local maxima for this function since it is a parabola opening upwards.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Function
Understanding a quadratic function is crucial when learning about maxima and minima. A quadratic function is any function that can be written in the form \( y = ax^2 + bx + c \). Here, "a", "b", and "c" are constants, with "a" not being zero. These functions graph into a parabolic shape, resembling a "U" or an upside-down "U", based on whether "a" is positive or negative.
- If "a" is positive, the parabola opens upwards, like open hands (\( y = x^2 \))
- If "a" is negative, it opens downwards, like closed hands (\( y = -x^2 \))
Derivative
The derivative of a function gives us information about its rate of change. To find where a function reaches its maximum or minimum, determining the derivative is an essential step. In calculus, the first derivative of a function \( y = f(x) \) is represented by \( \frac{dy}{dx} \).
In the case of our quadratic function \( y = 2x^2 - 8x + 6 \), the derivative was calculated to be \( \frac{dy}{dx} = 4x - 8 \).
In the case of our quadratic function \( y = 2x^2 - 8x + 6 \), the derivative was calculated to be \( \frac{dy}{dx} = 4x - 8 \).
- The derivative tells us how steep the slope of the tangent line to the curve is at any given point.
- It is used to find where the function changes direction, which typically happens where the derivative is equal to zero.
- This point is known as a critical point, which leads us to our next section.
Second Derivative
Once we have found the critical points of a function by solving the equation where its derivative is zero, we use the second derivative test to determine the nature of these points. The second derivative, \( \frac{d^2y}{dx^2} \), indicates the concavity of the function, telling us whether the function is opening upwards or downwards at a particular point.
For the quadratic function \( y = 2x^2 - 8x + 6 \), the second derivative is simply \( 4 \), which is a constant and positive.
For the quadratic function \( y = 2x^2 - 8x + 6 \), the second derivative is simply \( 4 \), which is a constant and positive.
- A positive second derivative (\( \frac{d^2y}{dx^2} > 0 \)) indicates that the graph is concave up, resembling a cup, and thus, the critical point is a local minimum.
- Conversely, a negative second derivative (\( \frac{d^2y}{dx^2} < 0 \)) indicates concave down, suggesting a local maximum.
Critical Points
The critical points of a function are where its derivative equals zero or is undefined. They are significant because they are potential locations for local maxima, minima, or saddle points. For the given equation \( y = 2x^2 - 8x + 6 \), setting the first derivative \( 4x - 8 \) to zero gave us the critical point \( x = 2 \).
- These points are identified by solving \( \frac{dy}{dx} = 0 \).
- They allow us to discuss the behavior of the function around these values.
- In combination with the second derivative, they help us classify the nature of each critical point.
