Find the absolute maximum and minimum values of f(x)=4x+4x^2, if any, over the interval (−∞,+∞).?

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  • 2 months ago
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    f '(x) = 4 + 8x

    Stationary points occur when f '(x) = 0

    i.e. 4 + 8x = 0

    so, x = -1/2

    f''(x) = 8...i.e. minimum as f''(x) > 0

    f(-1/2) = 4(-1/2) + 4(-1/2)² => -1

    Hence, miimum at (-1/2, -1)

    As x --> ∞, f(x) --> ∞

    As x --> -∞, f(x) --> ∞

    :)>

  • 2 months ago

    Differentiate and equate to zero . 

    f(x) = 4x + 4x^2 

    f'(x) = 4 + 8x = 0 

    8x = -4 

    x = -1/2 

    When x = -1/2 

    f(-1/2) = 4(-1/2) + 4(-1/2)^2 = -2 + 1 = -1 

    Hence the turning point is ( -1/2 , -1)

    We differentiate again. If the result is positive(+)/Negative(-) then it is  minimum/Maximum. 

    Hence f'(x) = 4 + 8x 

    f''(x) = (+)8 ; result is positive so it is a minimum at the turning point. 

    NB When x =  +/−∞ Then y = +∞ at a maximum. 

  • 2 months ago

    Hint: for x very very large and positive, x^2 is super big.

    Hint: for x very very large and negative, x^2 is super big.

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