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]
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
yea i already found the critical points
What are they?
x = 1, x = -1.3660, x = 0.36603
Now for the first part of the problem ie a)
Since he said in limits [-1,1], then what are the critical values that lie in the given limit?
so i substitute the critical point into the original problem right?
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
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
got it? when x lies [-2,2] that means \[-2\le x \le2\] but when x lies in (-2,2) that means \[-2 < x < 2\]
ok
Now, coming to the problem a)
Now say what are the critical points in the given region [-1,1] out of three you got?
1
and also 0.36603 right ?
cause its in limit [-1,1]
yes
Now while calculating absolute extrema, you should substitute the critical points in given limit and also upper limit and lower limit in the function
So, here for the a) problem you have to substitute x =1, x= 0.36603, x = -1
ok s (1) ^4 - 3(1)^2 + 2(1) + 1 = 1
and and for the second of i got -3.8480
and for the third i got 1.61592
so what do i do next?
what value did you get for x = 0.36603 ?
i got 1.61592
i got 1.3480
ohh ok
now which values of x has most positive and most negative values?
1.3480 is the maximum and -3.8480 is the minimum
for which value of x did you get -3.8480?
-1.3660
I said that you have to consider the critical values that are present in the given limit , not outside values!!!!!!!!!!
ok
-1.3660 is out of limit [-1,1] right?
yea
So, now tell me what are the maxima and minima that you got in the given interval
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
ok
tell me the values of extrema!
f(2) = 9 and f(-3) = -113 right?
f(-3) =49
ok
for the second problem . they are the extremas
ok
so the minimum is (-1.3660,-3.8480) and the max is (-3,49) ?
yes!
okay thanks
Your Welcome!
Join our real-time social learning platform and learn together with your friends!