# y=(x^2–x–2)^4: Find any turning points, determine their nature, and sketch the curveWatch

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Thread starter 2 weeks ago
#1
I differentiated using the chain rule to get 4(2x–1)(x^2–x–2)^3
I found the stationary points at x=-1, 0.5 and 2 by equating to zero

How do I determine their nature? Am I expected to find the second derivative using the product rule, despite the textbook not covering that method until a few pages later, or is there another way?

How do I sketch this curve? I am familiar with looking at repeated roots to determine the shape for functions of x up to the fourth power, but how can I apply this to higher powers like in this question?

Thank you
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2 weeks ago
#2
(Original post by TheOceanOwl)
I differentiated using the chain rule to get 4(2x–1)(x^2–x–2)^3
I found the stationary points at x=-1, 0.5 and 2 by equating to zero

How do I determine their nature? Am I expected to find the second derivative using the product rule, despite the textbook not covering that method until a few pages later, or is there another way?

How do I sketch this curve? I am familiar with looking at repeated roots to determine the shape for functions of x up to the fourth power, but how can I apply this to higher powers like in this question?

Thank you
The originial function is a quartic (of a quadratic x^2).
The quartic power makes everything non-negative, so the function must have two minimums which correspond to the quadratic = 0.
There must be a maximum in between. These are your stationary points.
Last edited by mqb2766; 2 weeks ago
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Thread starter 1 week ago
#3
(Original post by mqb2766)
The originial function is a quartic (of a quadratic x^2).
The quartic power makes everything non-negative, so the function must have two minimums which correspond to the quadratic = 0.
There must be a maximum in between. These are your stationary points.
Okay, thanks, I think I understand. So how would I verify stationary inflection at x=-1 and maximum at x=0 for y=(x³–x²+2)³ without the second derivative?
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1 week ago
#4
(Original post by TheOceanOwl)
Okay, thanks, I think I understand. So how would I verify stationary inflection at x=-1 and maximum at x=0 for y=(x³–x²+2)³ without the second derivative?
Why not just expand fully, then differentiate twice? No need to product rule then.
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1 week ago
#5
(Original post by TheOceanOwl)
Okay, thanks, I think I understand. So how would I verify stationary inflection at x=-1 and maximum at x=0 for y=(x³–x²+2)³ without the second derivative?
IMO, these questions are wanting you to think about how if you know how f(x) behaves in terms of turning points, and how g(x) behaves in terms of turning points, then this gives you a pretty good idea about how f(g(x)) behaves.

In this case g(x) = x^3 - x^2 + 2, and f(x) = x^3. When x = -1, g(x) = 0, so when we think about f(g(x)) we're at the inflection point. At the same time, f(x) is increasing, so we're travelling through the inflection point from left to right. So f(g(x)) is increasing, has a stationary point, and then increases again. So x = -1 is a stationary inflection point.

When x = 0, g(x) = 2, so when we thing about f(g(x)) we're at a point in the curve that's increasing. So if g(x) is increasing f(g(x)) is increasing and if g(x) is decreasing, f(g(x)) is decreasing. When we look at g' and g'' we see that x = 0 is a maximum point, which means g(x) is increasing to the left and decreasing from the right and this is enough to deduce f(g(x)) has a maximum.
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1 week ago
#6
(Original post by TheOceanOwl)
Okay, thanks, I think I understand. So how would I verify stationary inflection at x=-1 and maximum at x=0 for y=(x³–x²+2)³ without the second derivative?
Just reason about it.
The quartic is non-negative and only equal to zero when the quadratic is zero. They must be minimum values. Between two minimum (valleys) you must have a maximum (hill).

Note that at the minimums, the 2nd derivative is probably zero.
Last edited by mqb2766; 1 week ago
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1 week ago
#7
(Original post by mqb2766)
Just reason about it.
The quartic is non-negative and only equal to zero when the quadratic is zero. They must be minimum values. Between two minimum (valleys) you must have a maximum (hill).

Note that at the minimums, the 2nd derivative is probably zero.
I think the OP has moved on to a different question now. (i.e. y = (x^3-x^2+2)^3 is NOT a typo of the previous question).
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Thread starter 1 week ago
#8
(Original post by DFranklin)
IMO, these questions are wanting you to think about how if you know how f(x) behaves in terms of turning points, and how g(x) behaves in terms of turning points, then this gives you a pretty good idea about how f(g(x)) behaves.

In this case g(x) = x^3 - x^2 + 2, and f(x) = x^3. When x = -1, g(x) = 0, so when we think about f(g(x)) we're at the inflection point. At the same time, f(x) is increasing, so we're travelling through the inflection point from left to right. So f(g(x)) is increasing, has a stationary point, and then increases again. So x = -1 is a stationary inflection point.

When x = 0, g(x) = 2, so when we thing about f(g(x)) we're at a point in the curve that's increasing. So if g(x) is increasing f(g(x)) is increasing and if g(x) is decreasing, f(g(x)) is decreasing. When we look at g' and g'' we see that x = 0 is a maximum point, which means g(x) is increasing to the left and decreasing from the right and this is enough to deduce f(g(x)) has a maximum.
Thanks! So showing the stationary point plus whether it's increasing or decreasing either side is enough?
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