$$ \begin{array}{|c||c|c|c|} \hline x & 0 & 1 & 2 \\ \hline f(x) & 1 & k & 3 \\ \hline \end{array} $$

The function \(f\) is continuous on the closed interval \([0,2]\) and has the values that are given in the table above. The equation \(f(x)=-1\) must have a least two solutions in the interval \([0,2]\) if \(k=\)

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It may be difficult to analyze this problem and use the intermediate value theorem. Sketching a graph of the points and solution equation gives us a much more intuitive view:

For \(f\) to be continous and have two solutions where \(f(x)=-1\), we would need something like this:

The point, \((1,k)\), must have a \(k\)-value less than \(-1\).