How can I find the max/min of this function?

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In summary: The critical point is at x≈0.311 and x≈8.021, and you can use the first or second derivative test to determine the type of critical point. In summary, the critical points of the function f(x)=(-5x^2+3x)/(2x^2-5) are x≈0.311 and x≈8.021. The type of critical point can be determined using the first or second derivative test.
  • #1
randy17
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Homework Statement


Find the critical point of f(x)=(-5x^2+3x)/(2x^2-5), and show whether this point is a max/min.

Homework Equations



f(x)=(-5x^2+3x)/(2x^2-5)

The Attempt at a Solution


When I tried solving for the derivative of; f(x)=(-5x^2+3x)/(2x^2-5), I got (-6x^2+50x-15)/2x^2-5)^2.

then I set it equal to 0 and I got 0=-6x^2+50x-15

Now from here do I use the quadratic formula to solve for x? I tried that and I am getting x=(25+sqrt535)/6 and x=(25-sqrt535)/6. Have I done it right so far? When I try to plug in the x-values into the original eq to get y my calc says error... What do I do now?
 
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  • #2
randy17 said:
What do I do now?

Try again. Everything looks OK.
 
  • #3
Make sure you are using the right quadratic formula. Remember, it starts out with '-b', and your answers don't have the correct value for '-b'.
 
  • #4
The op's quadratic answer is correct.* If you're plugging in the exact formula into the calculator, I would simply instead just take an approximation of your answers (ie .311 and 8.02) and plug that in and then proceed with determining if the values are a max or a min.

*(The -b is negated by the 2a, where a is negative.)
 
Last edited:
  • #5
randy17 said:

Homework Statement


Find the critical point of f(x)=(-5x^2+3x)/(2x^2-5), and show whether this point is a max/min.

Homework Equations



f(x)=(-5x^2+3x)/(2x^2-5)

The Attempt at a Solution


When I tried solving for the derivative of; f(x)=(-5x^2+3x)/(2x^2-5), I got (-6x^2+50x-15)/2x^2-5)^2.

then I set it equal to 0 and I got 0=-6x^2+50x-15

Now from here do I use the quadratic formula to solve for x? I tried that and I am getting x=(25+sqrt535)/6 and x=(25-sqrt535)/6. Have I done it right so far? When I try to plug in the x-values into the original eq to get y my calc says error... What do I do now?

Your answers are correct; it sounds like you need to buy a new calculator.
 

What is the definition of a maximum or minimum for a function?

A maximum for a function is the highest point on the graph, where the value of the function is the largest. A minimum is the lowest point on the graph, where the value of the function is the smallest.

How can I find the max/min of a function algebraically?

To find the maximum or minimum of a function algebraically, you can take the derivative of the function and set it equal to zero. Then, solve for the variable to find the critical points. Plug these critical points into the original function to determine the maximum or minimum value.

What is the difference between a local and a global maximum/minimum?

A local maximum or minimum is a point on the graph where the value of the function is the highest or lowest within a specific interval. A global maximum or minimum is the highest or lowest point on the entire graph.

Can a function have more than one maximum or minimum?

Yes, a function can have multiple maximum or minimum points. These are called local extrema and can occur at points where the derivative is equal to zero or does not exist.

Are there any shortcuts or tricks to finding the max/min of a function?

There are some common techniques for finding the maximum or minimum of a function, such as using the first or second derivative tests, or graphing the function to visually determine the maximum or minimum points. However, these methods may not always work for every function and it is important to understand the underlying concepts and techniques for finding extrema in order to accurately solve the problem.

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