| Given g(x) | = | [f(2 f(x) + 2)]2 |
| ⇒ g'(x) | = | 2 f(2 f(x) + 2)).f '(2 f(x) + 2).2f '(x) |
| Substituting x | = | 0, |
| ⇒ g'(0) | = | 2f(2 f(0) + 2)).f '(2f(0) + 2).2f '(0) |
| = | 2f(0).f '(0).2f '(0) (∵ f(0) = –1 and f '(0) = 1) | |
| = | 2(–1)(1).2(1) = – 4 |

Now f(x) |
= | ex = 1 |
f(x) |
= | e–x = 1 |

f(x) = f(0) = 0
xp
= 0
exists.
exists
exists
xp – 1
does not exist