- #1
juantheron
- 247
- 1
No. of Real solution of ##2^x-x^2 = 1##
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
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