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Relative minima & maxima review

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.2 (EK)
Review how we use differential calculus to find relative extremum (minimum and maximum) points.

How do I find relative minimum & maximum points with differential calculus?

A relative maximum point is a point where the function changes direction from increasing to decreasing (making that point a "peak" in the graph).
Similarly, a relative minimum point is a point where the function changes direction from decreasing to increasing (making that point a "bottom" in the graph).
Supposing you already know how to find increasing & decreasing intervals of a function, finding relative extremum points involves one more step: finding the points where the function changes direction.
Want to learn more about relative extrema and differential calculus? Check out this video.

Example

Let's find the relative extremum points of f, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 3, x, squared, minus, 9, x, plus, 7. First, we differentiate f:
f, prime, left parenthesis, x, right parenthesis, equals, 3, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 1, right parenthesis
Our critical points are x, equals, minus, 3 and x, equals, 1.
Let's evaluate f, prime at each interval to see if it's positive or negative on that interval.
Intervalx-valuef, prime, left parenthesis, x, right parenthesisVerdict
x, is less than, minus, 3x, equals, minus, 4f, prime, left parenthesis, minus, 4, right parenthesis, equals, 15, is greater than, 0f is increasing. \nearrow
minus, 3, is less than, x, is less than, 1x, equals, 0f, prime, left parenthesis, 0, right parenthesis, equals, minus, 9, is less than, 0f is decreasing. \searrow
x, is greater than, 1x, equals, 2f, prime, left parenthesis, 2, right parenthesis, equals, 15, is greater than, 0f is increasing. \nearrow
Now let's look at the critical points:
xBeforeAfterVerdict
minus, 3\nearrow\searrowMaximum
1\searrow\nearrowMinimum
In conclusion, the function has a maximum point at x, equals, minus, 3 and a minimum point at x, equals, 1.

Check your understanding

Problem 1
h, left parenthesis, x, right parenthesis, equals, minus, x, cubed, plus, 3, x, squared, minus, 4
For what value of x does h have a relative maximum ?
Choose 1 answer:
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • blobby green style avatar for user D. Ashley Nelson
    (x−3)^2(1)+(x−2)[2(x−3)]
    = (x-3)^2+((x-2)2)(x-3)
    = (x-3)^2+(2x-4)(x-3)
    .... then I am lost. how do I get from here to ...
    ​=(x−3)(x−3+2x−4) ??

    Thank you
    (12 votes)
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  • starky seedling style avatar for user JK Jones
    I don understand how to do it without a graph.
    (1 vote)
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    • aqualine ultimate style avatar for user Ken
      If you don't want to hear me take a long winded explanation of why it works then just skip.
      If you think of maximum, it's like a hill. Say you can only climb the hill form the left to the right. If you're beginning to climb it, it's sloping up. But once you reach the top, it will start sloping down. And Since it must go from sloping up to sloping down in a continuous fashion the top point must have a slope of 0. (vice versa for minimum)

      First you take the derivative of an arbitrary function f(x). So now you have f'(x). Find all the x values for which f'(x) = 0 and list them down. So say the function f'(x) is 0 at the points x1,x2 and x3. Now test the points in between the points and if it goes from + to 0 to - then its a maximum and if it goes from - to 0 to + its a minimum
      (24 votes)
  • blobby green style avatar for user teconomides
    what if there are 2 variables F(x,y) = x^3 +y^2 -xy +x
    (2 votes)
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  • leafers tree style avatar for user R Haq
    so if f'(x) goes from + to - at x=c then f(c) is an maximum but how do i know if it is a relative or absolute max
    (2 votes)
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  • starky seed style avatar for user Jess Irish
    How do you solve for x^2/(x^2 -1)
    (2 votes)
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  • leafers sapling style avatar for user Gemma
    Values less than one are increasing (becoming less negative) and not decreasing, therefore there should only be one point with that is a retaliative minimum, that point being when x=1.
    (2 votes)
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  • blobby green style avatar for user santra313
    ) f(x) = 12x5 – 45x4 + 40x3 + 5. Find the value of x for which the curve shows relative maxima & relative minima
    (1 vote)
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    • mr pink red style avatar for user David Lee
      This is really simple if you watched videos. Find the first derivative of a function f(x) and find the critical numbers. Then, find the second derivative of a function f(x) and put the critical numbers. If the value is negative, the function has relative maxima at that point, if the value is positive, the function has relative maxima at that point. This is the Second Derivative Test. However, if you get 0, you have to use the First Derivative Test. Just find the first derivative of a function f(x) and critical numbers. Then, divide the domain (all real numbers) by the critical numbers. For example, if the critical numbers are -1, 4, 5, you should get 4 different domains which are (-∞, -1), (-1, 4), (4, 5), (5, ∞). Then, input any number inside each domain. If the value is negative and the value of next domain is positive, it has relative minima at that point and vice versa. For example, if you got a negative value in the interval (-1, 4) and positive value in the interval (4, 5), the function has relative minima at point 4.

      Hope this helps! If you have any questions or need help, please ask! :)
      (2 votes)
  • blobby green style avatar for user Zaina Tarafder
    When we're doing the number line, what if it is negative before and after the point? Does that mean it is just continuously decreasing?
    (1 vote)
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  • blobby green style avatar for user Elyssa Laurel - P6
    given the function f(x) = x^3+ax^2+bx+12, when x=3 there is a relative minimum value of -15. solve for a and b
    (1 vote)
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    • leaf green style avatar for user kubleeka
      You know that plugging in x=3 will give a value of -15, so that gives you an equation in a and b.

      If you take the derivative of f and plug in x=3, you know it will give 0, since you have a relative minimum. This gives you a second equation in a and b. Solve the system for a and b.
      (2 votes)
  • blobby green style avatar for user Doubra Anoruse
    How do you find the y value for the given points
    (1 vote)
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