**problem**: Show that there is at least one value of \({x}\) for which \({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}\)

**solution**: \({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}\)

\({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1}\) is a continuous function. Now if we can chose \({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1}\) takes both negative & positive numbers then the \({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 = 0}\) have a solution.

For \({\displaystyle{x = {\frac{1}{64}}}}\), \({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 < 0}\) & for \({x = 64}\), \({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 > 0}\).

So by intermediate value theorem we can say that \({\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 = 0}\) has a solution for some real \({x}\).

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