Solution of exp equation

  • Thread starter juantheron
  • Start date
  • #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
 

Answers and Replies

  • #2
UltrafastPED
Science Advisor
Gold Member
1,914
216
Note that f'''(x) is always positive. This tells you something about f''(x).

Draw a simple graph of these derivatives.
 

Suggested for: Solution of exp equation

Replies
3
Views
113
Replies
10
Views
467
Replies
3
Views
390
Replies
4
Views
555
Replies
3
Views
388
Replies
2
Views
369
  • Last Post
2
Replies
68
Views
3K
  • Last Post
Replies
2
Views
76
Replies
5
Views
216
Top