Find maxima/minima of polynomials

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The discussion focuses on finding the maxima and minima of the polynomial function p(x) = -(x+1)(x-1)^2. Participants discuss using the product rule for differentiation and simplifying the expression to find critical points. The critical values identified are x = 1 and x = -1/3, which are potential local extrema. However, it is noted that the function does not have global maxima or minima without restrictions on x, as it approaches infinity in both directions. The conversation emphasizes the importance of understanding local versus global extrema in polynomial functions.
Orson
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Homework Statement


find maxima/minima of following equation.

Homework Equations


-(x+1)(x-1)^2

The Attempt at a Solution


(-x-1)(x-1)^2

Using product rule, we obtain,
-1(x-1)^2+(-x-1)*2(x-1)
I don't know where to go from here. The software's factoring I had never seen before.
 
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Orson said:

Homework Statement


find maxima/minima of following equation.

Homework Equations


-(x+1)(x-1)^2

The Attempt at a Solution


(-x-1)(x-1)^2

Using product rule, we obtain,
-1(x-1)^2+(-x-1)*2(x-1)
I don't know where to go from here. The software's factoring I had never seen before.
Factor out an ##(x-1)## and simplify it. No software needed.
 
LCKurtz said:
Factor out an ##(x-1)## and simplify it. No software needed.
How do i do that with the 2 and the -1?

and the (-x-1)
 
Write ##(x-1)(...)##, put what is left in the other parentheses, and simplify it.
 
LCKurtz said:
Write ##(x-1)(...)##, put what is left in the other parentheses, and simplify it.
-1(x-1)^2+(-x-1)*2(x-1)
=(x-1)((x-1)(-1)+2(-x-1))
ok. now what?
 
Orson said:
-1(x-1)^2+(-x-1)*2(x-1)
=(x-1)((x-1)(-1)+2(-x-1))
ok. now what?
Simplify the second factor.
 
LCKurtz said:
Simplify the second factor.
(x-1)(-3x-1)=0
x=1, -1/3
 
Orson said:
(x-1)(-3x-1)=0
x=1, -1/3
OK, you have the critical values of ##x##. Hopefully you can take it from here. It's sack time for me.
 
LCKurtz said:
OK, you have the critical values of ##x##. Hopefully you can take it from here. It's sack time for me.
I can. thank you very much. have a good night.
 
  • #10
Orson said:

Homework Statement


find maxima/minima of following equation.

Homework Equations


-(x+1)(x-1)^2

The Attempt at a Solution


(-x-1)(x-1)^2

Using product rule, we obtain,
-1(x-1)^2+(-x-1)*2(x-1)
I don't know where to go from here. The software's factoring I had never seen before.
Forget the software. A diff. rule you should have already seen is the constant multiple rule: ##\frac d {dx}(k \cdot f(x)) = k \frac d {dx} f(x)##
Treat the minus sign as being a multiplier of -1, so ##\frac d {dx}[-(x + 1)(x - 1)^2] = -1 \cdot \frac d{dx}[(x + 1)(x - 1)^2]##
If you can use the product rule on the part in the brackets, and bring along the factor of -1, you should be able to work through this.
 
  • #11
Orson said:

Homework Statement


find maxima/minima of following equation.

Homework Equations


-(x+1)(x-1)^2

The Attempt at a Solution


(-x-1)(x-1)^2

Using product rule, we obtain,
-1(x-1)^2+(-x-1)*2(x-1)
I don't know where to go from here. The software's factoring I had never seen before.

Have you written the complete problem statement, with no clarifying words missing? I ask, because strictly speaking your function ##p(x) = -(x+1)(x-1)^2## does not have a maximum or a minimum when we place no restrictions on ##x##. For any number ##N > 0##, no matter how large, we can easily find values ##x_1## and ##x_2## giving ##p(x_1) > N## and ##p(x_2) < -N##. (In other words, the maximum of ##p(x)## is ##+\infty## and its minimum is ##-\infty##.) However, ##p(x)## does have local maxima and minima within restricted regions of ##x##
 
  • #12
Ray Vickson said:
Have you written the complete problem statement, with no clarifying words missing? I ask, because strictly speaking your function ##p(x) = -(x+1)(x-1)^2## does not have a maximum or a minimum when we place no restrictions on ##x##. For any number ##N > 0##, no matter how large, we can easily find values ##x_1## and ##x_2## giving ##p(x_1) > N## and ##p(x_2) < -N##. (In other words, the maximum of ##p(x)## is ##+\infty## and its minimum is ##-\infty##.) However, ##p(x)## does have local maxima and minima within restricted regions of ##x##
Ray Vickson said:
Have you written the complete problem statement, with no clarifying words missing? I ask, because strictly speaking your function ##p(x) = -(x+1)(x-1)^2## does not have a maximum or a minimum when we place no restrictions on ##x##. For any number ##N > 0##, no matter how large, we can easily find values ##x_1## and ##x_2## giving ##p(x_1) > N## and ##p(x_2) < -N##. (In other words, the maximum of ##p(x)## is ##+\infty## and its minimum is ##-\infty##.) However, ##p(x)## does have local maxima and minima within restricted regions of ##x##
It was multiple choice on khan academy. for which values of x is the function a local minimum or maximum. i can't remember which. But -1/3 was the correct answer per the software (s-word again)
 

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