<p>Answer #1: First consider the circumstances of a point of inflection on f(x), and in particular what that would cause on f '(x) and f ‘’(x). A point of inflection on f(x), in which the concavity changes from positive to negative or vice versa, reflects a change in slope from positive to negative or vice versa on f '(x) and an actual value change from positive to negative or vice versa on f ‘’(x).</p>
<p>For example, if there is a point of inflection at 2.3 on a function g(x) where the function switches from concave-up to concave-down, that means that on the function g '(x) there is a maximum (positive slope –> 0 slope –> negative slope), where the maximum occurs at 2.3. On the function g ‘’(x), the values of g(x) actually change from positive (before x = 2.3) to negative (after x = 2.3), and at 2.3 x would have a value of 0.</p>
<p>Keeping this in mind, let us address the problem above. Let f be the function with derivative defined by f '(x) = sin (x^3) on the interval, -1.8 is less than x which is less than 1.8. How many points of inflection does the graph of f have on this interval?</p>
<p>Since we know we are going to need f ‘’(x), let us first calculate that using the chain rule:</p>
<p>u = x^3.
y = f '(x) = sin(u).</p>
<p>f ‘’(x) = dy/dx = dy/du x du/dx.</p>
<p>f ‘’(x) = cos(u) x (3x^2) = (3x^2)cos(x^3).</p>
<p>Now that we have f ‘’(x), we need to set it equal to zero to find the possible points of inflection. Note that not every point where f ‘’(x) = 0 is an inflection point – an inflection point only occurs where on either side of the x where f ‘’(x) = 0 the values of f ‘’(x) switch from positive to negative or vice versa.</p>
<p>f ‘’(x) = (3x^2)cos(x^3) = 0</p>
<p>(3x^2) = 0 OR cos(x^3) = 0
x = 0 OR (x^3 = pi/2 or 3pi/2)
x = 0 OR (x = (pi/2)^(1/3) or x = (3pi/2)^(1/3))</p>
<p>Note that since this function’s interval is (-1.8,1.8), you need to factor in equivalent negative possible x values. Thus the x values where f ‘’(x) = 0 on that interval are:</p>
<p>x = -(3pi/2)^(1/3)
x = -(pi/2)^(1/3)
x = 0
x = (pi/2)^(1/3)
x = (3pi/2)^(1/3)</p>
<p>The next part is most easily done with a number line, but basically you want to break the number line into intervals from one possible x to another. You should get something like:</p>
<p>^
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-1.68
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-1.16
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|
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0
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|
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+1.16
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+1.68
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v</p>
<p>The values on the vertical number line above mark the intervals. Now remember what we said in the beginning: a flip from concave up to concave down or vice versa in f(x) is the same as a flip from positive to negative or vice versa in f ‘’(x). Test values within each interval and plug into the f ‘’(x) function to see whether the results are positive or negative. You should get the following order of positive-negative intervals:</p>
<p>[+,-,+,+,-,+]</p>
<p>Now the question is how many switches are there. If you count the number of times that list switches from positive to negative or negative to positive, you should count 4, which is the answer to your question.</p>
<p>I hope I explained this reasonably well – it’s a little confusing sometimes >.<</p>
<p>good luck~!</p>