Calculus (Help please)

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Find all the critical numbers for the following function.
F(x) = (16 - x^2)^(3/5)

  • Calculus (Help please) -

    f' = (3/5)(16-x^2)^(-2/5)(-2x)
    That is undefined when x = 4 or x = -4 and is zero when x = 0

  • Calculus (Help please) -

    f'(x) = (3/5)(16 - x^2)^(-2/5) (-2x)
    = 0 for max/min of f(x)
    -2x = 0
    x = 0 , then y = 16^(3/5)

    find f''(x) using the product rule
    set that equal to zero to find any points of inflection

    x-intercepts, let y = 0
    (16-x^2)^(3/5) = 0
    x^2 = 16
    x = ± 4

    y-intercept, let x = 0
    y = 16^(3/5)

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