# Find maxima/minima of polynomials

Orson

## Homework Statement

find maxima/minima of following equation.

-(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.

Science Advisor
Homework Helper
Gold Member

## Homework Statement

find maxima/minima of following equation.

-(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.

Orson
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)

Science Advisor
Homework Helper
Gold Member
Write ##(x-1)(...)##, put what is left in the other parentheses, and simplify it.

Orson
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?

Science Advisor
Homework Helper
Gold Member
-1(x-1)^2+(-x-1)*2(x-1)
=(x-1)((x-1)(-1)+2(-x-1))
ok. now what?
Simplify the second factor.

Orson
Simplify the second factor.
(x-1)(-3x-1)=0
x=1, -1/3

Science Advisor
Homework Helper
Gold Member
(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.

Orson
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.

Mentor

## Homework Statement

find maxima/minima of following equation.

-(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.

Science Advisor
Homework Helper
Dearly Missed

## Homework Statement

find maxima/minima of following equation.

-(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##

Orson
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)