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

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 $-1$.