Factor out an ##(x-1)## and simplify it. No software needed.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.
How do i do that with the 2 and the -1?LCKurtz said:Factor out an ##(x-1)## and simplify it. No software needed.
-1(x-1)^2+(-x-1)*2(x-1)LCKurtz said:Write ##(x-1)(...)##, put what is left in the other parentheses, and simplify it.
Simplify the second factor.Orson said:-1(x-1)^2+(-x-1)*2(x-1)
=(x-1)((x-1)(-1)+2(-x-1))
ok. now what?
(x-1)(-3x-1)=0LCKurtz said:Simplify the second factor.
OK, you have the critical values of ##x##. Hopefully you can take it from here. It's sack time for me.Orson said:(x-1)(-3x-1)=0
x=1, -1/3
I can. thank you very much. have a good night.LCKurtz said:OK, you have the critical values of ##x##. Hopefully you can take it from here. It's sack time for me.
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)##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.
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.
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)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##
A maximum/minimum of a polynomial is a point on the graph of the polynomial where the function either reaches its highest or lowest value. It is also known as a turning point or critical point.
To find the maximum/minimum of a polynomial, you can use the derivative of the polynomial. Set the derivative equal to 0 and solve for the x-values. These x-values will be the critical points, which can then be plugged into the original polynomial to find the corresponding y-values for the maximum/minimum points.
Yes, a polynomial can have multiple maximum/minimum points. This can occur when the polynomial has multiple local extrema, meaning it changes from increasing to decreasing or vice versa at several points on the graph.
A local extremum is a maximum/minimum that occurs at a specific point on the graph, while a global extremum is the absolute maximum/minimum for the entire graph. To determine if a maximum/minimum is a global or local extremum, you can look at the behavior of the polynomial before and after the critical point. If the polynomial is increasing before the critical point and decreasing after, it is a local maximum. If the polynomial is decreasing before and increasing after, it is a local minimum. To determine if it is a global extremum, you would need to compare it to all other critical points on the graph.
Yes, calculus can be used to find the maximum/minimum of any polynomial. However, for more complex polynomials, the process may be more involved and require advanced calculus techniques such as the second derivative test or optimization methods.