Show that the function x^4+3z+1 have exactly one zero in the interval [-2,-1]
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The function f(x)=x^4+3x+1 is decreasing on the interval [-2,-1]. To show this, we take the derivative.
f'(x)=4x^3+3, which is less than zero whenever
4x^3<-3, which is the same asx^3<-3/4, or equivalently
x<(-3/4)^(1/3). Since (-3/4)^(1/3) is greater than -1, the derivative is negative on the whole interval, which means the function is decreasing and thus can have at most one zero there.
Now, f(-2)=(-2)^4+3(-2)+1=11, which is greater than 0, and f(-1)=(-1)^4+3(-1)+1=-1, less than
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answer will I think upper once is correct
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