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
[tex]x=\frac{4}{3}[/tex]
g)
f(1)=1-3+4=2
[tex]\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[/tex]
[tex]\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}[/tex]
[tex]\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[/tex]
sau stim ca
[tex]\lim_{x \to \inft1} \frac{f(x)-f(1)}{x-1} = f'(1)=-3[/tex]
h)
[tex]\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}[/tex]
[tex]\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}=[/tex]
[tex]\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}[/tex]
[tex]\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[/tex]
sau stim ca
[tex]\lim_{x \to \inft-1} \frac{g(x)-g(-1)}{x+1} =g'(-1)=21[/tex]
