f(x)=1/sqrt(1−|x|)+1/x^4 I was asked to show though the intermediate value theorem that f(x) has at least one solution when f(x)=314. I found that f(x) is continuous when −1<x<0 and 0<x<1. The problem I'm having is that I thought that the theorem only worked for closed intervalls. Any tips are greatly appreciated!

Colton Gregory

Colton Gregory

Open question

2022-08-19

f ( x ) = 1 1 | x | + 1 x 4
I was asked to show though the intermediate value theorem that f ( x ) has at least one solution when f ( x ) = 314. I found that f ( x ) is continuous when 1 < x < 0 and 0 < x < 1. The problem I'm having is that I thought that the theorem only worked for closed intervalls. Any tips are greatly appreciated!

Answer & Explanation

Kendrick Mendez

Kendrick Mendez

Beginner2022-08-20Added 9 answers

When x = 0.000001 , f is huge. When x = 0.5, f is small. Somewhere in the middle is 314.

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