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Math Help - Can anyone help me figure this out. Critical points.

  1. #1
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    Can anyone help me figure this out. Critical points.

    (a) Let a > 0 be a constant. Find all critical points of f (x) = x + a√x.

    This is my work so far

    f'(x)=1+(a/2)(x^(-1/2))
    0=1+(a/2)(x^(-1/2))
    -1 = (a/2)(1/sqrt(x))
    -1=a/(2sqrt(x))
    (2)(-1)=(a/(2sqrt(x)))(2)
    -2=a/sqrtx
    (a)-1/2=sqrtx/a (a)
    -a/2=sqrtx
    (-a/2)^2=(sqrtx)^2
    a/4=x

    I calculated the Critical point as x=a/4

    (b) Use derivatives to show that f is increasing and its graph is concave down for all x > 0.

    Not sure how to do this part.
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  2. #2
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    Quote Originally Posted by dptrimble View Post
    (a) Let a > 0 be a constant. Find all critical points of f (x) = x + a√x.

    This is my work so far

    f'(x)=1+(a/2)(x^(-1/2))
    0=1+(a/2)(x^(-1/2))
    -1 = (a/2)(1/sqrt(x))
    -1=a/(2sqrt(x))
    (2)(-1)=(a/(2sqrt(x)))(2)
    -2=a/sqrtx
    (a)-1/2=sqrtx/a (a)
    -a/2=sqrtx
    (-a/2)^2=(sqrtx)^2
    a/4=x

    I calculated the Critical point as x=a/4

    (b) Use derivatives to show that f is increasing and its graph is concave down for all x > 0.

    Not sure how to do this part.

    The function's derivative is 1+\frac{a}{2\sqrt{x}}=1+\frac{a}{2}x^{-\frac{1}{2}}

    Thus, f'(x)>0\, \forall{x} and thus it is increasing.
    Now evaluate the second derivative and prove it is always negative.

    Tonio
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