Exercise · ZW-INTERMEDIATE-VALUE-THEOREM-008 · ★★★ Advanced
Let \(f(x)=1/x\) on \([-1,1]\). Can the theorem be used to find \(c\) with \(f(c)=0\)?
Choose one option.
A Yes, because \(f\) is differentiable on \((-1,1)\) B No, \(f\) is not continuous on \([-1,1]\) because it is undefined at \(0\) C Yes, because \(f(-1)=-1\) and \(f(1)=1\) have opposite signs D Yes, because \(f\) is odd
Show answer B. No, \(f\) is not continuous on \([-1,1]\) because it is undefined at \(0\)
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Explanation Although \(f(-1)\) and \(f(1)\) have opposite signs, \(f\) is not continuous at \(0\), so the theorem does not apply. Indeed \(1/x\) never equals \(0\).
If \(f\) is continuous on \([a,b]\) and \(N\) lies between \(f(a)\) and \(f(b)\), there is some \(c\in(a,b)\) with \(f(c)=N\).