calculus

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Find the extreme values of the function for the given intervals.

2x^3 – 21x^2 + 72x

a) [0,5]

b) [0,4]

I have both the minimum values for both intervals which is 0. But I am unable to find the maximum. please help!!!

  • calculus -

    You have probably found two extrema where f'(x)=0.
    To identify whether the extremum is a maximum or minimum, use the second derivative test.
    If the second derivative is positive, the extremum is a minimum. If the second derivative is negative, the extremum is a maximum.

    Here the extrema are at x=3 and x=4.

    The second derivative, f"(x)=12x-42,
    so for example, f2(3)=-6, so x=3 is a maximum.

    Use your calculator to plot the graph of the function to help you solve these problems. After some time, you will be able to visualize the graph from the equation, even without the calculator.

  • calculus -

    I tried 3 for both (a and b) as a maximum and didn't work....

  • calculus -

    It is important to know that extreme values include local maxima, local minima and end-points of the given interval.

    The question requires the extreme values of the function, so it is the value of the function that counts.

    To find extreme values, we calculate the critical points (where f'(0)=0 and where f'(0) does not exist) as well as the value of the function at end-points.

    Given
    f(x)=2x^3 – 21x^2 + 72x

    1. We calculate the points where f'(x)=0, and we have determined that f(3)=81 is a local maximum and f(4)=80 is a local minimum.

    2. For [0,4], the values of the function at end-points are:
    f(0)=zero, and f(4)=80.
    So the list of values to consider are:
    (0,zero)
    (3,81)
    (4,80)
    We conclude that the extreme minimum is zero, and the extreme maximum is 81.

    3. For [0,5], the values of the function at end-points are f(0)=zero, and f(5)=85.
    So the list of values to consider for extreme values are:
    (0,zero)
    (3,81)
    (4,80)
    (5,85)
    We choose zero as the extreme minimum, and 85 as the extreme maximum.

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