a) f(-2) = -1; f(-1) = 0; f(0) = 1; f(2) = 3

g(-1) = 0,5; g(-2) = 2; g(0) = 0

b) f(x) = 2 ⇒ x = 1

g(x) = 2 ⇒ x = 2 hoặc x = -2

Bài 1:

Để \(F\left(x\right)=G\left(x\right)\) thì \(3x^2-8x+4=3x+4\)

\(\Leftrightarrow3x^2-11x=0\)

\(\Leftrightarrow x\left(3x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{11}{3}\end{matrix}\right.\)

a, f(-1) =2.(-1)+1

=-2+1

=-1

g(-1)=1-2.-1

=1-(-2)

=3

2a,

f(0)=0² +7.0+12

=12

g(4)=4² -12

=4

1bta có :

f(g(x))=f(g(-1))

g(-1) =3=> f(3)

f(3)=2.3+1=7

g(f(x))=g(f(-1))

f(-1)=-1=> g(-1)

g(-1)=1-2.(-1)=3

Chọn C.

Đặt  u   =   G ( x ) d v   =   f ( x ) d x ⇒ d u   =   G ( x ) ' d x   =   g ( x )   d x v   =   ∫ f ( x ) d x   =   F ( x )

Suy ra: I =  G ( x ) F ( x ) 2 0   - ∫ 0 2 F ( x ) g ( x ) d x  

= G(2)F(2) – G(0)F(0) – 3 = 1 – 0 – 3 = -2.

a: f(a)=g(a)

=>5a-3=-1/2a+1

=>5,5a=4

=>\(a=\dfrac{4}{5.5}=\dfrac{8}{11}\)

b: f(b-2)=g(2b+4)

=>\(5\left(b-2\right)-3=-\dfrac{1}{2}\left(2b+4\right)+1\)

=>\(5b-13=-b-2+1=-b-1\)

=>6b=12

=>b=2

f(a) = g(a)

⇔ 5a - 3 = -a/2 + 1

⇔ 5a + a/2 = 1 + 3

⇔ 11a/2 = 4

⇔ 11a = 8

⇔ a = 8/11

Vậy a = 8/11 thì f(a) = g(a)

b) f(b - 2) = g(2b + 4)

⇔ 5.(b - 2) - 3 = -(2b + 4)/2 + 1

⇔ 5b - 10 - 3 = -b - 2 + 1

⇔ 5b + b = 1 + 13

⇔ 6b = 14

⇔ b = 7/3

Vậy b = 7/3 thì f(b - 2) = g(2b + 4)

Để hàm số có đạo hàm tại x=0 phải thỏa mãn 2 điều kiện, đó là hàm số liên tục tại x=0 và có đạo hàm bên trái bằng đạo hàm bên phải

Để hàm số liên tục tại x=0 \(\Leftrightarrow\lim\limits_{x\rightarrow0^+}=\lim\limits_{x\rightarrow0^-}=f\left(0\right)\Leftrightarrow2=2\left(tm\right)\)

\(f'\left(0^+\right)=\lim\limits_{x\rightarrow0^+}\dfrac{f\left(x\right)-f\left(0\right)}{x-0}=\lim\limits_{x\rightarrow0^+}\dfrac{mx^2+2x+2-2}{x}=\lim\limits_{x\rightarrow0^+}\dfrac{x\left(mx+2\right)}{x}=2\)

\(f'\left(0^-\right)=\lim\limits_{x\rightarrow0^-}\dfrac{f\left(x\right)-f\left(0\right)}{x-0}=\lim\limits_{x\rightarrow0^-}\dfrac{nx+2-2}{x}=n\)

\(\Rightarrow\left\{{}\begin{matrix}m\in R\\n=2\end{matrix}\right.\)

\(f\left(0^+\right)=f\left(0^-\right)\Leftrightarrow n=2\)

a: \(F\left(3\right)=3\left(3-2\right)=3\cdot1=3\)

\(\left[F\left(\dfrac{2}{3}\right)\right]^2=\left[\dfrac{2}{3}\cdot\left(\dfrac{2}{3}-2\right)\right]^2\)

\(=\left[\dfrac{2}{3}\cdot\dfrac{-4}{3}\right]^2=\left(-\dfrac{8}{9}\right)^2=\dfrac{64}{81}\)

\(G\left(-\dfrac{1}{2}\right)=-\left(-\dfrac{1}{2}\right)+6=6+\dfrac{1}{2}=\dfrac{13}{2}\)

b: F(x)=0

=>x(x-2)=0

=>\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c: F(a)=G(a)

=>\(a\left(a-2\right)=-a+6\)

=>\(a^2-2a+a-6=0\)

=>\(a^2-a-6=0\)

=>(a-3)(a+2)=0

=>\(\left[{}\begin{matrix}a-3=0\\a+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-2\end{matrix}\right.\)