<p>This is a question from CB all subject tests prep.</p>
<p>Can anyone please explain how to solve this problem without a calculator?</p>
<p>The function f is defined by f(x)=x^4-4x^2+x+1 for =5<=x<=5.
In which of the following intervals does the minimum value of f occur?</p>
<p>(a)-5<x<-3
(b)-3<x<-1
(c)-1<x<1
(d)1<x<3
(e)3<x<5</p>
<p>f’(x)=4x^3-8x+1
For extrema to occur, f’(x)=0
Try different values of x,
f’(1)=-3<0, f’(2)=17>0
For f’(x) increases from negative to positive in the interval of 1<x<2,
there must be a 1<x=x0<2 so that f’(x0)=0
And it indicates that f(x) attains minimum when 1<x<2 for the graph changes from negative slope to positive in this interval.
Hence the answer should be (d).</p>
<p>there are two minims in these intervals. one is between -3 and -1, another is between 1 and 3. from those to the most minimum value will lye between -3 and -1. so i think the correct answer is b. i tried using my graphing calculator. very sorry if my answer is wrong!!! try it using graphing calculator.</p>