Rolle's Theorem and The Mean Value Thm Indeterminate Forms and L'Hospital's Rule Antiderivative # Rolle's Theorem and The Mean Value Thm# Rolle's ThmSuppose f f f satifies that
f : [ a , b ] → R f:[a,b] \to \mathbb{R} f : [ a , b ] → R is continuous.f : ( a , b ) → R f:(a,b) \to \mathbb{R} f : ( a , b ) → R is differentiable.f ( a ) = f ( b ) f(a) = f(b) f ( a ) = f ( b ) then ∃ c ∈ ( a , b ) \exists c \in (a,b) ∃ c ∈ ( a , b ) such that f ′ ( c ) = 0 f'(c) = 0 f ′ ( c ) = 0 # The Mean Value ThmSuppose f f f satifies that
f : [ a , b ] → R f:[a,b] \to \mathbb{R} f : [ a , b ] → R is continuous.f : ( a , b ) → R f:(a,b) \to \mathbb{R} f : ( a , b ) → R is differentiable. then ∃ c ∈ ( a , b ) \exists c \in (a,b) ∃ c ∈ ( a , b ) such that f ′ ( c ) = f ( b ) − f ( a ) b − a f'(c) = \frac{f(b)-f(a)}{b-a} f ′ ( c ) = b − a f ( b ) − f ( a ) .Consider lim x → a f ( x ) g ( x ) \lim_{x \to a}\frac{f(x)}{g(x)} lim x → a g ( x ) f ( x ) :
若當x → a x \to a x → a 時,同時有 f ( x ) → 0 f(x) \to 0 f ( x ) → 0 及 g ( x ) → 0 g(x) \to 0 g ( x ) → 0 , 則lim x → a f ( x ) g ( x ) \lim_{x \to a}\frac{f(x)}{g(x)} lim x → a g ( x ) f ( x ) 稱為0 0 \frac{0}{0} 0 0 不定型 若當x → a x \to a x → a 時,同時有 f ( x ) → ∞ f(x) \to \infty f ( x ) → ∞ 及 g ( x ) → ∞ g(x) \to \infty g ( x ) → ∞ , 則lim x → a f ( x ) g ( x ) \lim_{x \to a}\frac{f(x)}{g(x)} lim x → a g ( x ) f ( x ) 稱為∞ ∞ \frac{\infty}{\infty} ∞ ∞ 不定型 # L'Hospital's RuleSuppose f f f and g g g are differentiable and g ′ ( x ) ≠ 0 g'(x) \neq 0 g ′ ( x ) = 0 on ( a , b ) (a,b) ( a , b ) containing c c c . Suppose that lim x → a f ( x ) = 0 \lim_{x \to a}f(x) = 0 lim x → a f ( x ) = 0 and lim x → a g ( x ) = 0 \lim_{x \to a}g(x) = 0 lim x → a g ( x ) = 0 or lim x → a f ( x ) = ± ∞ \lim_{x \to a}f(x) = \pm \infty lim x → a f ( x ) = ± ∞ and lim x → a g ( x ) = ± ∞ \lim_{x \to a}g(x) = \pm \infty lim x → a g ( x ) = ± ∞ Then lim x → a f ( x ) g ( x ) = lim x → a f ′ ( x ) g ′ ( x ) \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} lim x → a g ( x ) f ( x ) = lim x → a g ′ ( x ) f ′ ( x ) if the limit on the right hand side exists(or is ± ∞ \pm \infty ± ∞ )
# Antiderivative# DefineA function F F F is called an antiderivative of f f f on an open interval I I I if F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) for all x ∈ I x \in I x ∈ I .
# Examplefunctions antiderivatives functions antiderivatives c f ( x ) cf(x) c f ( x ) c F ( x ) + k cF(x)+k c F ( x ) + k s i n x sinx s i n x − c o s x + k -cosx+k − c o s x + k f ( x ) + g ( x ) f(x)+g(x) f ( x ) + g ( x ) F ( x ) + G ( x ) + k F(x)+G(x)+k F ( x ) + G ( x ) + k s e c 2 x sec^2x s e c 2 x t a n x + k tanx+k t a n x + k x n ( n ≠ − 1 ) x^n(n \neq -1) x n ( n = − 1 ) 1 n + 1 x n + 1 + k \frac{1}{n+1}x^{n+1}+k n + 1 1 x n + 1 + k t a n x s e c x tanxsecx t a n x s e c x s e c x + k secx+k s e c x + k 1 x \frac{1}{x} x 1 .l n ∥ x ∥ + k ln\|x\|+k l n ∥ x ∥ + k 1 1 − x 2 \frac{1}{\sqrt{1-x^2}} 1 − x 2 1 .s i n − 1 x + k sin^{-1}x+k s i n − 1 x + k e x e^x e x e x + k e^x+k e x + k 1 1 + x 2 \frac{1}{1+x^2} 1 + x 2 1 .t a n − 1 x + k tan^{-1}x+k t a n − 1 x + k c o s x cosx c o s x s i n x + k sinx+k s i n x + k − 1 1 − x 2 \frac{-1}{\sqrt{1-x^2}} 1 − x 2 − 1 .c o s − 1 x + k cos^{-1}x+k c o s − 1 x + k
# Reference蘇承芳老師 - 微積分甲(一)109 學年度 - Calculus (I) Academic Year 109