• Support PF! Buy your school textbooks, materials and every day products Here!

Solution of exp equation

  • Thread starter juantheron
  • Start date
  • #1
21
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
1,912
216
Note that f'''(x) is always positive. This tells you something about f''(x).

Draw a simple graph of these derivatives.
 

Related Threads on Solution of exp equation

Replies
1
Views
2K
Replies
4
Views
3K
Replies
9
Views
937
  • Last Post
Replies
8
Views
704
  • Last Post
Replies
6
Views
767
Replies
3
Views
923
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
1
Views
1K
Top