Find maxima/minima of polynomials

  • #1
Orson
67
5

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.
 

Answers and Replies

  • #2
LCKurtz
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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.
Factor out an ##(x-1)## and simplify it. No software needed.
 
  • #3
Orson
67
5
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)
 
  • #4
LCKurtz
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Write ##(x-1)(...)##, put what is left in the other parentheses, and simplify it.
 
  • #5
Orson
67
5
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?
 
  • #6
LCKurtz
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-1(x-1)^2+(-x-1)*2(x-1)
=(x-1)((x-1)(-1)+2(-x-1))
ok. now what?
Simplify the second factor.
 
  • #7
Orson
67
5
Simplify the second factor.
(x-1)(-3x-1)=0
x=1, -1/3
 
  • #8
LCKurtz
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(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.
 
  • #9
Orson
67
5
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
36,851
8,880

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
Ray Vickson
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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.

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
Orson
67
5
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##
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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