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}.