Question

Ma poate ajuta cineva cu tot exercitiul va rog
Am nevoie

Answer 1

f(x)=x³-3x²+4

g(x)=x³-5x²+8x-4

a)

f'(x)=(x³-3x²+4)'=3x²-6x

g'(x)=(x³-5x²+8x-4)'=3x²-10x+8

f'(x)-g'(x)=3x²-6x-(3x²-10x+8)=4x-8

b)

f'(x)=0

3x²-6x=0

3x(x-2)=0

3x=0⇒x=0

x-2=0⇒x=2

c)

f'(1)+g'(-1)=-3+21=18

f'(1)=3×1-6×1=-3

g'(-1)=3×(-1)²-10×(-1)+8=3+10+8=21

d)

f"(x)-g"(x)=

f"(x)=(3x²-6x)'=6x-6

g"(x)=(3x²-10x+8)'=6x-10

f"(x)-g"(x)=6x-6-6x+10=4

e)

f"(x)=0

6x-6=0

6x=6

x=1

f)

f"(x)+g"(x)=0

6x-6+6x-10=0

12x-16=0

12x=16 |:4

3x=4

$x=\frac{4}{3}$

g)

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

$\lim_{x \to \inft1} \frac{f(x)-f(1)}{x-1} = \lim_{x \to \inft1} \frac{x^3-3x^2+4-2}{x-1} =\lim_{x \to \inft1} \frac{x^3-3x^2+2}{x-1}=\frac{0}{0} \ forma\ nedeterminata$

$\lim_{x \to \inft1} \frac{x^3-3x^2+2}{x-1}=\lim_{x \to \inft1} \frac{x^3-x^2-2x^2+2}{x-1}=\lim_{x \to \inft1} \frac{x^2(x-1)-2(x^2-1)}{x-1}=\lim_{x \to \inft1} \frac{x^2(x-1)-2(x-1)(x+1)}{x-1}$

$\lim_{x \to \inft1} \frac{x^2(x-1)-2(x-1)(x+1)}{x-1}=\lim_{x \to \inft1} \frac{(x-1)(x^2-2x-2)}{x-1}=\lim_{x \to \inft1}x^2-2x-2=1-2-2=-3$

sau stim ca

$\lim_{x \to \inft1} \frac{f(x)-f(1)}{x-1} = f'(1)=-3$

h)

$\lim_{x \to \inft-1} \frac{g(x)-g(-1)}{x+1} = \lim_{x \to \inft-1} \frac{x^3-5x^2+8x-4+18}{x+1}$

$\lim_{x \to \inft-1} \frac{x^3-5x^2+8x+14}{x+1}= \lim_{x \to \inft-1} \frac{x^3+1-5x^2+5+8x+8}{x+1}=$

$\lim_{x \to \inft-1} \frac{(x+1)(x^2-x+1)-5(x-1)(x+1)+8(x+1)}{x+1}=\lim_{x \to \inft-1} \frac{(x+1)(x^2-x+1-5x+5+8)}{x+1}$

$\lim_{x \to \inft-1} \frac{(x+1)(x^2-x+1-5x+5+8)}{x+1}=\lim_{x \to \inft-1} x^2-x+1-5x+5+8=\lim_{x \to \inft-1} x^2-6x+14=21$

sau stim ca

$\lim_{x \to \inft-1} \frac{g(x)-g(-1)}{x+1} =g'(-1)=21$