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Solution (Given as):: Clearly ##x = 0## and ##x = 1## are solution of Given equation.

Formal; define ##f(x) = 2^x - x^2 - 1##, so ##f'(x) = 2^x\ln 2 - 2x##, ##f''(x) = 2^x\ln^2 2 - 2##,

##f'''(x) = 2^x\ln^3 2 > 0##. Study the variation, in order to estimate the values of the roots and

critical points

Now I Did not Under How can we check Nature of ##f(x)## using ##f^{'''}(x)##

Please Help me

Thanked