Finding Extremum Points of f(x): A Study

[zorro]

Homework Statement

Let f(x) =(x-1)^p.(x-2)^q where p,q>1. Each critical point of f(x) is a point of extremum when - (Options are given)

The Attempt at a Solution

I got the critical points as 1 and 2.
I don't know what do I do next. I found the second derivative but I think its of no use.
Someone enlighten me with their thougts.
Thanks.

 

[ptolema]

plug the critical points into f''. if f''(critical point)>0, critical point is a minimum. if f''(critical point)<0, critical point is a max. if f''(critical point)=0, use f' to tell where f changes fro increasing to decreasing.

this stuff is http://www.mathwords.com/f/first_derivative_test.htm"

 

[zorro]

I should have written the options to make it more clear.
a) p =3, q=4
b) p =5, q=7
c) p=2, q=3
d) p=2, q=4

How do you find the values of p and q?

 

[lanedance]

Ok, so as p and q>1, the critical points will be at x=1,2 now as suggested calculate the 2nd derivative and see what effect the given values of p and q have

 

[zorro]

I don't want to plug in each value of p and q.
How do I find out the values of p and q without looking at the options?
I found out the second derivative. It becomes 0 at the critical points.

 

[lanedance]

so show what the 2nd derivative is

 

[zorro]

second derivative is

f²(x) = f(x) (-p/(x-1)² - q/(x-2)²) + f¹(x)(p/(x-1) + q/(x-2))

 

[lanedance]

i don't think that's quite right, to start i get
f(x) =(x-1)^p.(x-2)^q
f&#039;(x) =p(x-1)^{p-1}.(x-2)^q + q(x-1)^p(x-2)^{q-1}

 

[zorro]

I rechecked it. Its correct. Take (x-1)^p and (x-2)^q common from the first derivative and then differentiate. You will get same.

 

[lanedance]

the following forms may be useful
f&#039;(x) =(x-1)^{p-1}(x-2)^{q-1}(p(x-2)+ q(x-1))
f&#039;&#039;(x) = (x-1)^{p-2}(x-2)^{q-2}(p(p-1)(x-2)^2+ pq(x-1))(x-2)+q(q-1)(x-1)^2)

if f''(x) is zero, it does not tell you whether it is an extrememum or not and you will need to examine higher derivatives

 

[zorro]

Yes I know that and it will require a lot of time.
There must be some other way to solve this because tht time allotted for this question is 2 mins.

 

[Mark44]

You can do this problem with taking any derivatives.

a)p = 3, q = 4
f(x) = (x - 1)³(x - 2)⁴
With this combination of values, the graph of f rises out of the third quadrant, crosses the x-axis at x = 1, dips back down to the x-axis at x = 2, and rises up into the first quadrant.

The zeros are (obviously) at x = 1 and x = 2.
Near x = 1, the graph of f looks like y = x³ shifted to the right by 1 unit, so x = 1 cannot be an extremum for these value of p and q.

Near x = 2, the graph of f looks like y = x⁴ shifted to the right by 2 units. There is a local extremum (minimum) at x = 2.

And so on for the other three sets of values...

 

[zorro]

Thanks a lot Mark44.
I got the answer as d)