Question 1079044: Please help me solve this
Consider the function f(x) = x^3 + px + q.
(a) Determine the values of p and q if f(x) has a stationary point at (−2,3).
(b) Show, using calculus, that there is a second stationary point at (2,−29), and classify both
stationary points.
(c) Determine f′′(x) and hence show that there is a non-stationary point of inflection and determine its coordinates.
(d) For what values of k would the equation f(x) = k have 3 distinct solutions. Give reasons for your answer.
(Hint: Sketch the graph of y = f(x).)
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! A stationary point is where the first derivative is defined and equal to zero
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f'(x) = 3x^2 + p
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we know x = -2 at the stationary point (-2,3)
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1) 3(-2)^2 + p = 0
p = -12
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we know f(-2) = 3 since (-2,3) is a point on the graph of f(x)
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2) 3 = (-2)^3 -12(-2) + q
3 = -8 +24 + q
q = -13
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f(x) = x^3 -12x -13
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a. p = -12, q = -13
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f'(x) = 3x^2 -12
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3x^2 -12 = 0
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3x^2 = 12
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x^2 = 4
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x = 2 and x = -2
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b. for x = 2
f(2) = 2^3 -12(2) -13 = -29
stationary point at (2,-29)
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:
f'(x) = 3x^2 -12
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An inflection point is a point on a curve where the curve changes sign.
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we take the second derivative, which is
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f''(x) = 6x
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our candidate x value is x = 0
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Note that if x is negative then 6x is negative and if x is positive then 6x is positive, therefore the point with x = 0 is an inflection point
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f(0) = 0^3 -12(0) -13 = -13
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c. the point of inflection is (0,-13)
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the graph of f(x) = x^3 -12x -13 looks like
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we want values of k, such that f(x) crosses the x axis 3 times
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d. looking at the graph, this is where f(x) = 0, that is, k = 0
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