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Math Help - Functions

  1. #1
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    Functions

    hi,

    Q1.

    f(x)=(x-2)(x+1)^2

    a) find the min and max values of f(x):

    b) A point of inflection is a point where the graph changes direction of concavity, i.e. a point where f''(x)=0. Find the inflection point(s) of f, and determine where the function is concave up and concave down.

    ok..

    for a),
    i presumed that if y'=0 then it is either the min/max value,
    so using product rule, i got 0=2(x+1)(x-2)
    i.e. x=-1 or x=2
    so with those x values, the function will be at its min = 0,
    i also presumed that the max value is infinity...


    for b), well im stuck on that


    could anyone help me on both questions?

    thanks,

    moon
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by Moon Hoplite View Post
    hi,

    Q1.

    f(x)=(x-2)(x+1)^2

    a) find the min and max values of f(x):

    b) A point of inflection is a point where the graph changes direction of concavity, i.e. a point where f''(x)=0. Find the inflection point(s) of f, and determine where the function is concave up and concave down.

    ok..

    for a),
    i presumed that if y'=0 then it is either the min/max value,
    so using product rule, i got 0=2(x+1)(x-2)
    i.e. x=-1 or x=2
    so with those x values, the function will be at its min = 0,
    i also presumed that the max value is infinity...


    for b), well im stuck on that


    could anyone help me on both questions?

    thanks,

    moon
    The product rule states that
    \frac{d}{dx} f(x) g(x) = f'(x) g(x) + g'(x) f(x)

    Thus, the correct derivative of the function f(x) = (x-2)(x+1)^2 is:
    (x+1)^2 + 2(x+1)(x-2) = (x+1)(x+1+2x-4) = 3(x-1)(x+1)

    Now you say \frac{d}{dx} f(x) = 3(x-1)(x+1) = 0 and solve for x, which gives you extreme points at x=-1,1.

    To find the points of inflection or to determine concavity, differentiate once again: \frac{d^2}{dx^2} = 3(x+1) + 3(x-1) = 6x = 0. Your only inflection point is at x=0.

    To verify whether x=-1,1 is a minimum or maximum, check the concavity at those points. So plug in x=-1 into \frac{d^2}{dx^2} = 6x to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for x=1.

    It helps to graph the function to verify.
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  3. #3
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    Quote Originally Posted by Last_Singularity View Post
    The product rule states that
    \frac{d}{dx} f(x) g(x) = f'(x) g(x) + g'(x) f(x)

    Thus, the correct derivative of the function f(x) = (x-2)(x+1)^2 is:
    (x+1)^2 + 2(x+1)(x-2) = (x+1)(x+1+2x-4) = 3(x-1)(x+1)

    Now you say \frac{d}{dx} f(x) = 3(x-1)(x+1) = 0 and solve for x, which gives you extreme points at x=-1,1.

    To find the points of inflection or to determine concavity, differentiate once again: \frac{d^2}{dx^2} = 3(x+1) + 3(x-1) = 6x = 0. Your only inflection point is at x=0.

    To verify whether x=-1,1 is a minimum or maximum, check the concavity at those points. So plug in x=-1 into \frac{d^2}{dx^2} = 6x to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for x=1.

    It helps to graph the function to verify.

    hmmn thanks, but im confused on question 1, so does y'=0 states the min or max x values? and if so is the max value at infinity?
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  4. #4
    Member Last_Singularity's Avatar
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    To verify whether x=-1,1 is a minimum or maximum, check the concavity at those points. So plug in x=-1 into \frac{d^2}{dx^2} = 6x to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for x=1.


    To verify whether x=-1,1 is a minimum or maximum, check the concavity at those points. So plug in x=-1 into \frac{d^2}{dx^2} = 6x to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for x=1.


    To verify whether x=-1,1 is a minimum or maximum, check the concavity at those points. So plug in x=-1 into \frac{d^2}{dx^2} = 6x to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for x=1.


    To verify whether x=-1,1 is a minimum or maximum, check the concavity at those points. So plug in x=-1 into \frac{d^2}{dx^2} = 6x to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for x=1.
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