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Concavity for f(x) = 9 (x - 5)^{2/3}

  1. Jan 14, 2014 #1
    I just have a question regarding the function f(x) = 9 (x - 5)^{2/3}. I did the all the stuff to find concavity and I got my answer that it's never concave up and concave down on (-∞,5)U(5,∞). But I don't know what exactly my prof is saying (ie. the initial answer to the question was wrong so I corrected it by saying: "The second derivative is -2/(x-5)^(4/3). So it is obvious that f"(5) does not exist and that at all other points the function is negative." He wrote back and said that my answer was right but for the wrong reason. He saying that "If f is concave down on (-inf, inf), then all tangents lie above the curve for all real numbers. But the tangent at x=6 intersects the curve for some x-value less than 0." I don't understand what he's trying to say, I mean isn't the answer simply just because the second derivative dne at x = 5 but negative everywhere else, you have to break up the interval for concavity at x = 5?
     
  2. jcsd
  3. Jan 14, 2014 #2

    Mark44

    Staff: Mentor

    I don't understand what your prof is saying, either. f'' is defined on two separate intervals, just as you say. On each of these intervals, the graph of f is concave down. I don't see how it's reasonable to say that "f is concave down on (-∞, ∞)..."

    Also, I disagree with what he said about the tangent at x = 6 intersecting the curve at some negative x value. Because of the 2/3 power, f(x) ≥ 0 for all real x. The graph of f is symmetric about the vertical line x = 5. For the tangent at (5, f(5)) to intersect the graph of f at some negative x value, the graph of f would have to lie below the x-axis for x < 0.

    I suspect that your prof is looking at a graph such as the one produced by wolframalpha, which is incorrect, due to the way that WA computes rational powers (it converts mr to er*ln(m)). Here's a link to the WA plot - http://www.wolframalpha.com/input/?i=y+=+9(x-5)^(2/3). Notice that the real-valued plot doesn't show anything to the left of x = 5, which is incorrect.
     
  4. Jan 14, 2014 #3
    He also said this: "f is not concave down on (-inf, inf) since the tangent line at (6,9) does not lie completely above the graph (or since the secant line joining (4,9) to (6,9) does not lie below the graph)." I don't know what idea he's trying to reference that I may be neglecting.

    Also, even though it doesn't show the other half of the graph, it is obvious that the tangent line does intersect the graph and some other point. Is he trying to say that the tangent line cannot touch any other part of the curve (so literally concave down means that the tangent line ACTUALLY lies about the curve and doesn't touch any other points? Because I think there is a tangent line on x^3 that intersects other points).
     
  5. Jan 14, 2014 #4
    x^3 isn't concave downward on the entire real line. Your professor is saying that functions that are concave downward at every point do have their tangent lines above the graph.
     
  6. Jan 14, 2014 #5
    Oh! So for example, if g(x) = -e^x, then the second derivative is -e^x which is always negative, so g is always concave down. I plotted it on Wolfram and all the tangent lines do not intersect the graph and all lie above the graph. Is this what he meant? I guess he was trying to determine it by looking at the graph instead of working out the algebra. But by using the second derivative and noticing that at x = 5, f''(5) does not exist, is this sufficient in showing that the function is concave down on (-∞,5)U(5,∞)? All I was trying to do was to correct the answer that the online assessment thing was giving.

    In addition, I said that f has to be differentiable on the interval you are joining the secant line, but it's not (at x = 5). But he said that "A function can be concave down on an interval I without being differentiable at every x in I. This is unusual for the type of functions we see in this course but it can happen." Can someone give an example of such a function?
     
  7. Jan 14, 2014 #6

    Mark44

    Staff: Mentor

    This question isn't about y = x3.
     
  8. Jan 14, 2014 #7
    I was trying to make a connection with a simple function but shortydeb explained that only if the function is concave down on (-infinity, infinity) then all the tangent lines lie above the curve and don't intersect the curve at all. In fact, this is exactly what my prof said in the email, I must've skimmed it too fast :(.
     
  9. Jan 14, 2014 #8

    Mark44

    Staff: Mentor

    Maybe y = -|x|. Strictly speaking this graph doesn't lie above its tangent lines. Your prof seems to be using a definition of concavity similar to this one (from Calculus and Analytic Geometry, Abraham Schwartz).
    The definition for concave down is similar.
     
  10. Jan 14, 2014 #9
    But for y = -|x|, y' is neither increasing nor decreasing, so it's neither concave up nor concave down.
     
  11. Jan 14, 2014 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The general definition of concavity is that ##f(x)## is concave (= concave down) on an interval ##I \subset \mathbb{R}## if
    [tex] f(\lambda x_1 + (1-\lambda) x_2) \geq \lambda f(x_1) + (1-\lambda) f(x_2) [/tex]
    for all ##x_1, x_2 \in I, 0 \leq \lambda \leq 1##. In other words, the line segment joining ##(x_1,f(x_1))## and ##(x_2, f(x_2))## lies (on or) below the graph ##y = f(x)## between ##x_1## and ##x_2##. For such ##f## that are twice differentiable we need ##f''(x) \leq 0 ## in ##I##, and at points ##x## where ##f'(x)## does not exist we need ##f'(x-) \geq f'(x+)##; in other words, if ##f'## has a jump discontinuity at ##x##, it jumps down as we pass from left to right through ##x##. So, ##f(x) = - |x|## is a perfectly good concave function. See, eg., http://en.wikipedia.org/wiki/Concave_function for more material.

    BTW: your function ##f(x) = 9 |x-5|^{2/3}## does not satisfy that jump condition at ##x = 5##, so is not concave on ##\mathbb{R}##, but it is concave on each part ##(-\infty,5)## and ##(5,\infty)## separately.
     
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