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Mathematics 18 Online
OpenStudy (anonymous):

Numerically estimate the absolute extrema of the given function on the indicated intervals. f(x) = x^4 - 3x^2 + 2x + 1 on a) [-1,1] b) [-3,2]

OpenStudy (anonymous):

derivate the function and find x values by equating the derivated equation to zero then you will have some x values. they are called critical values

OpenStudy (anonymous):

yea i already found the critical points

OpenStudy (anonymous):

What are they?

OpenStudy (anonymous):

x = 1, x = -1.3660, x = 0.36603

OpenStudy (anonymous):

Now for the first part of the problem ie a)

OpenStudy (anonymous):

Since he said in limits [-1,1], then what are the critical values that lie in the given limit?

OpenStudy (anonymous):

so i substitute the critical point into the original problem right?

OpenStudy (anonymous):

NOTE : the square brackets are inclusive i mean when you have given that x lies in limit [-2,2] x may take any value in between -2 and 2, x can also take the values 2 and -2

OpenStudy (anonymous):

but the open brackets "( )" are exclusive brackets i mean when you have given that x lies in limit [-2,2] x may take any value in between -2 and 2, but x can not take the values 2 and -2

OpenStudy (anonymous):

got it? when x lies [-2,2] that means \[-2\le x \le2\] but when x lies in (-2,2) that means \[-2 < x < 2\]

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Now, coming to the problem a)

OpenStudy (anonymous):

Now say what are the critical points in the given region [-1,1] out of three you got?

OpenStudy (anonymous):

1

OpenStudy (anonymous):

and also 0.36603 right ?

OpenStudy (anonymous):

cause its in limit [-1,1]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Now while calculating absolute extrema, you should substitute the critical points in given limit and also upper limit and lower limit in the function

OpenStudy (anonymous):

So, here for the a) problem you have to substitute x =1, x= 0.36603, x = -1

OpenStudy (anonymous):

ok s (1) ^4 - 3(1)^2 + 2(1) + 1 = 1

OpenStudy (anonymous):

and and for the second of i got -3.8480

OpenStudy (anonymous):

and for the third i got 1.61592

OpenStudy (anonymous):

so what do i do next?

OpenStudy (anonymous):

what value did you get for x = 0.36603 ?

OpenStudy (anonymous):

i got 1.61592

OpenStudy (anonymous):

i got 1.3480

OpenStudy (anonymous):

ohh ok

OpenStudy (anonymous):

now which values of x has most positive and most negative values?

OpenStudy (anonymous):

1.3480 is the maximum and -3.8480 is the minimum

OpenStudy (anonymous):

for which value of x did you get -3.8480?

OpenStudy (anonymous):

-1.3660

OpenStudy (anonymous):

I said that you have to consider the critical values that are present in the given limit , not outside values!!!!!!!!!!

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

-1.3660 is out of limit [-1,1] right?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

So, now tell me what are the maxima and minima that you got in the given interval

OpenStudy (anonymous):

while substituting the values of x for extrema, you have to substitute the upper limit and lower limit and also critical points in the interval

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

tell me the values of extrema!

OpenStudy (anonymous):

f(2) = 9 and f(-3) = -113 right?

OpenStudy (anonymous):

f(-3) =49

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

for the second problem . they are the extremas

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so the minimum is (-1.3660,-3.8480) and the max is (-3,49) ?

OpenStudy (anonymous):

yes!

OpenStudy (anonymous):

okay thanks

OpenStudy (anonymous):

Your Welcome!

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