Question

URGENT PENTRU BACC<<<<

Se considera functia f(x)=(x-5)(x-4)(x-3)(x-2)+1
aratati ca f'(5)=6

Answer 1

Răspuns:

Explicație pas cu pas:

(x - 5)(x - 4) = x^2 - 4x - 5x + 20 = x^2 - 9x + 20

(x - 3)(x - 2) = x^2 - 2x - 3x + 6 = x^2 - 5x + 6

(x^2 - 9x + 20)(x^2 - 5x + 6)

= x^4 - 5x^3 + 6x^2 - 9x^3 + 45x^2 - 54x + 20x^2 - 100x + 120

= x^4 - 14x^3 + 71x^2 - 154x + 120

f(x) = x^4 - 14x^3 + 71x^2 - 154x + 121

f'(x) = 4x^3 - 42x^2 + 142x - 154

f'(5) = 4*125 - 42*25 + 142*5 - 154

= 500 - 1050 + 710 - 154

= 6

Answer 2

Răspuns:

f(x) = (x-5)(x-4)(x-3)(x-2) + 1

derivata produsului de functii: (f₁·f₂·...fₙ)' = f₁'·f₂·...fₙ + f₁·f₂'·...fₙ + ... + f₁·f₂·...fₙ'

⇒ f'(x) = 1·(x-4)(x-3)(x-2) + (x-5)·1·(x-3)(x-2) + (x-5)(x-4)·1·(x-2) + (x-5)(x-4)(x-3)·1

f'(x) = (x-4)(x-3)(x-2) + (x-5)(x-3)(x-2) + (x-5)(x-4)(x-2) + (x-5)(x-4)(x-3)

f'(5) = (5-4)(5-3)(5-2) + 0 + 0 + 0 = 1·2·3 = 6

b)

$\lim_{n \to \infty} [\frac{f(n+1)-1}{f(n)-1}]^{n}=\\ = \lim_{n \to \infty} [\frac{(n-4)(n-3)(n-2)(n-1)+1-1}{(n-5)(n-4)(n-3)(n-2)+1-1}]^{n} =\\ = \lim_{n \to \infty} [\frac{(n-4)(n-3)(n-2)(n-1)}{(n-5)(n-4)(n-3)(n-2)}]^{n} =\\= \lim_{n \to \infty} [\frac{(n-1)}{(n-5)}]^{n}$

notez t = (n-1)/(n-5) = (1-1/n)/(1-5/n)

evident: n -> ∞ => t -> (1-0)/(1-0) = 1

t = (n-1)/(n-5) = t*n-5t = n-1

⇒ n = (5t-1)/(t-1) = (5t-5+5-1)/(t-1)

⇒ n = 5 + 4/(t-1)

$ln[ \frac{(n-1)}{(n-5)}]^{n} = n*ln [\frac{(n-1)}{(n-5)}]=( 5 + \frac{4}{t-1})ln(t) = 5ln(t)+\frac{4ln(t)}{t-1}$

$\lim_{n \to \infty} ln[ \frac{(n-1)}{(n-5)}]^{n} = \lim_{t \to 1}[ 5ln(t)+\frac{4ln(t)}{t-1}] =\\= \lim_{t \to 1}[ 5ln(t)]+ \lim_{t \to 1}[\frac{4ln(t)}{t-1}] =\\=5*\lim_{t \to 1}[ln(t)]+ +4*\lim_{t \to 1}[\frac{ln(t)}{t-1}] =\\= 5*0+4*1=\\=4$

$\lim_{n \to \infty} ln[ \frac{(n-1)}{(n-5)}]^{n} = 4 <=> ln[\lim_{n \to \infty} [ \frac{(n-1)}{(n-5)}]^{n}] = 4\\e^{ ln[\lim_{n \to \infty} [ \frac{(n-1)}{(n-5)}]^{n}] } = e^{4} \\\\=>\lim_{n \to \infty} [ \frac{(n-1)}{(n-5)}]^{n}= e^{4}$

obs: Mai sus am folosit:

$\lim_{t \to 1}[\frac{ln(t)}{t-1}] =1$

care se deduce astfel:

$\lim_{t \to 1}[\frac{ln(t)}{t-1}] = 0/0 =\lim_{t \to 1}[\frac{ln'(t)}{(t-1)'}] =\lim_{t \to 1}[\frac{1/t}{1}] =1$