If $f(x)=2x^2+8x+5$, Find The Relative Extrema.

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Introduction

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In calculus, finding the relative extrema of a function is a crucial concept that helps us understand the behavior of the function. The relative extrema of a function are the points where the function changes from increasing to decreasing or vice versa. In this article, we will find the relative extrema of the given function $f(x)=2x^2+8x+5$.

What are Relative Extrema?

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Relative extrema are the points on the graph of a function where the function changes from increasing to decreasing or vice versa. In other words, they are the points where the function has a local maximum or minimum value. The relative extrema can be found by taking the derivative of the function and setting it equal to zero.

Finding the Derivative

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To find the relative extrema of the function $f(x)=2x^2+8x+5$, we need to find the derivative of the function. The derivative of a function is a measure of how fast the function changes as its input changes. In this case, we can find the derivative of the function using the power rule of differentiation.

Power Rule of Differentiation

The Power Rule of Differentiation

The power rule of differentiation states that if $f(x)=x^n$, then $f'(x)=nx^{n-1}$. We can use this rule to find the derivative of the function $f(x)=2x^2+8x+5$.

Derivative of $f(x)=2x^2+8x+5$

Derivative of $f(x)=2x^2+8x+5$

Using the power rule of differentiation, we can find the derivative of the function $f(x)=2x^2+8x+5$ as follows:

$$f'(x)=\frac{d}{dx}(2x^2+8x+5)=4x+8$$

Setting the Derivative Equal to Zero

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To find the relative extrema of the function, we need to set the derivative equal to zero and solve for $x$. This will give us the critical points of the function.

Setting $f'(x)=0$

Setting $f'(x)=0$

Setting the derivative equal to zero, we get:

$$4x+8=0$$

Solving for $x$

Solving for $x$

Solving for $x$, we get:

$$x=-2$$

Second Derivative Test

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To determine whether the critical point is a local maximum or minimum, we need to use the second derivative test. The second derivative test states that if the second derivative of the function is positive at the critical point, then the function has a local minimum at that point. If the second derivative is negative, then the function has a local maximum at that point.

Second Derivative of $f(x)=2x^2+8x+5$

Second Derivative of $f(x)=2x^2+8x+5$

Taking the derivative of the derivative, we get:

$$f''(x)=\frac{d}{dx}(4x+8)=4$$

Applying the Second Derivative Test

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Now that we have the second derivative, we can apply the second derivative test to determine whether the critical point is a local maximum or minimum.

Second Derivative Test

Second Derivative Test

Since the second derivative is positive, we know that the function has a local minimum at the critical point.

Conclusion

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In this article, we found the relative extrema of the function $f(x)=2x^2+8x+5$ by taking the derivative of the function and setting it equal to zero. We then used the second derivative test to determine whether the critical point is a local maximum or minimum. The critical point is a local minimum, and the function has a minimum value at $x=-2$.

Relative Extrema of $f(x)=2x^2+8x+5$

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The relative extrema of the function $f(x)=2x^2+8x+5$ are:

• Local minimum: $x=-2$, $f(-2)=5$

Graph of $f(x)=2x^2+8x+5$

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The graph of the function $f(x)=2x^2+8x+5$ is a parabola that opens upward. The graph has a local minimum at $x=-2$.

Graph of $f(x)=2x^2+8x+5$

Graph of $f(x)=2x^2+8x+5$

The graph of the function $f(x)=2x^2+8x+5$ is shown below:

f(x) = 2x^2 + 8x + 5

Final Answer

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The final answer is $\boxed{-2}$.

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Introduction

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In our previous article, we found the relative extrema of the function $f(x)=2x^2+8x+5$ by taking the derivative of the function and setting it equal to zero. We then used the second derivative test to determine whether the critical point is a local maximum or minimum. In this article, we will answer some frequently asked questions about relative extrema.

Q&A

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Q: What is the definition of relative extrema?

A: Relative extrema are the points on the graph of a function where the function changes from increasing to decreasing or vice versa. In other words, they are the points where the function has a local maximum or minimum value.

Q: How do I find the relative extrema of a function?

A: To find the relative extrema of a function, you need to take the derivative of the function and set it equal to zero. This will give you the critical points of the function. Then, you need to use the second derivative test to determine whether the critical point is a local maximum or minimum.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine whether a critical point is a local maximum or minimum. If the second derivative of the function is positive at the critical point, then the function has a local minimum at that point. If the second derivative is negative, then the function has a local maximum at that point.

Q: Can a function have more than one relative extrema?

A: Yes, a function can have more than one relative extrema. For example, the function $f(x)=x^3-6x^2+9x+2$ has three relative extrema.

Q: How do I graph a function with relative extrema?

A: To graph a function with relative extrema, you need to plot the function and identify the critical points. Then, you need to use the second derivative test to determine whether the critical point is a local maximum or minimum. Finally, you can plot the local maximum or minimum points on the graph.

Q: What is the importance of relative extrema in real-world applications?

A: Relative extrema are important in real-world applications because they help us understand the behavior of a function. For example, in economics, relative extrema can help us understand the maximum or minimum profit of a company. In physics, relative extrema can help us understand the maximum or minimum velocity of an object.

Examples

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Example 1: Find the relative extrema of the function $f(x)=x^2-4x+3$

To find the relative extrema of the function $f(x)=x^2-4x+3$, we need to take the derivative of the function and set it equal to zero.

$$f'(x)=\frac{d}{dx}(x^2-4x+3)=2x-4$$

Setting the derivative equal to zero, we get:

$$2x-4=0$$

Solving for $x$, we get:

$$x=2$$

Then, we need to use the second derivative test to determine whether the critical point is a local maximum or minimum.

$$f''(x)=\frac{d}{dx}(2x-4)=2$$

Since the second derivative is positive, we know that the function has a local minimum at the critical point.

Example 2: Find the relative extrema of the function $f(x)=x^3-6x^2+9x+2$

To find the relative extrema of the function $f(x)=x^3-6x^2+9x+2$, we need to take the derivative of the function and set it equal to zero.

$$f'(x)=\frac{d}{dx}(x^3-6x^2+9x+2)=3x^2-12x+9$$

Setting the derivative equal to zero, we get:

$$3x^2-12x+9=0$$

Solving for $x$, we get:

$$x=1, x=3$$

Then, we need to use the second derivative test to determine whether the critical point is a local maximum or minimum.

$$f''(x)=\frac{d}{dx}(3x^2-12x+9)=6x-12$$

Since the second derivative is positive at $x=1$ and negative at $x=3$, we know that the function has a local minimum at $x=1$ and a local maximum at $x=3$.

Conclusion

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In this article, we answered some frequently asked questions about relative extrema. We also provided examples of how to find the relative extrema of a function using the second derivative test. We hope that this article has helped you understand the concept of relative extrema and how to apply it in real-world applications.

Final Answer

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The final answer is $\boxed{1, 3}$.