\(a,f\left(-2\right)=\dfrac{3}{4}\left(-2\right)=-\dfrac{3}{2}\\ f\left(0\right)=\dfrac{3}{4}\cdot0=0\\ f\left(1\right)=\dfrac{3}{4}\cdot1=\dfrac{3}{4}\\ b,g\left(-2\right)=\dfrac{3}{4}\left(-2\right)+3=-\dfrac{3}{2}+3=\dfrac{3}{2}\\ g\left(0\right)=\dfrac{3}{4}\cdot0+3=3\\ g\left(1\right)=\dfrac{3}{4}\cdot1+3=\dfrac{15}{4}\)

\(a.\)

Ta có : \(y=f\left(x\right)=4x^2-3\)

\(\Rightarrow f\left(-2\right)=4.\left(-2\right)^2-3=13\)

a) f(-2)=4*2²-3=13

b)f(-2)=(-2)

b) f(x)=x <=>4x^2_ 3= 4x^{2 _} x - 3= 0 <=>4x² - 4x+ 3x - 3= 0 <=>4x (x + 1)=0 <=> (x -1) (4x +3)= 0 <=> x=1 hoặc x= \(\frac{-3}{4}\)

a) thay f(-2) vào hàm số ta có :

y=f(-2)=(-4).(-2)+3=11

thay f(-1) vào hàm số ta có :

y=f(-1)=(-4).(-1)+3=7

thay f(0) vào hàm số ta có :

y=f(0)=-4.0+3=-1

thay f(-1/2) vào hàm số ta có :

y=f(-1/2)=(-4).(-1/2)+3=5

thay f(1/2) vào hàm số ta có :

y=f(-1/2)=(-4).1/2+3=1

b)

f(x)=-1 <=> -4x+3=-1 => x=1

f(x)=-3 <=> -4x+3=-3 => x=3/2

f(x)=7 <=> -4x+3=7 => x=-1

Bạn ơi, f(0)= -4.0 + 3 =3 mà!

ukm đúng rồi bạn , mình quên

\(f\left(1\right)=4\cdot1-5=-1\)

\(f\left(3\right)=4\cdot3-5=7\)

a) Ta có: 

\(f\left(-2\right)=\left|3\cdot-2-1\right|=\left|-6-1\right|=\left|-7\right|=7\) 

\(f\left(2\right)=\left|3\cdot2-1\right|=\left|6-1\right|=5\)

\(f\left(-\dfrac{1}{4}\right)=\left|3\cdot-\dfrac{1}{4}-1\right|=\left|-\dfrac{3}{4}-1\right|=\left|-\dfrac{7}{4}\right|=\dfrac{7}{4}\) 

b) Ta có: 

\(f\left(x\right)=10\)

\(\Rightarrow\left|3x-1\right|=10\)

Với \(x\ge\dfrac{1}{3}\Rightarrow3x-1=10\)

\(\Rightarrow3x=11\Rightarrow x=\dfrac{11}{3}\left(tm\right)\)

Với \(x< \dfrac{1}{3}\Rightarrow3x-1=-10\)

\(\Rightarrow3x=-9\Rightarrow x=-3\left(tm\right)\) 

_______

\(f\left(x\right)=-3\)

\(\Rightarrow\left|3x-1\right|=-3\)

Mà: \(\left|3x-1\right|\ge0\forall x\) và \(-3< 0\)

\(\Rightarrow\left|3x-1\right|=-3\) (vô lý)

\(\Rightarrow\) không có x thỏa mãn 

Ta có: \(y=f\left(x\right)=4x-3\)

\(f\left(1\right)=4.1-3=1\)

\(f\left(2\right)=4.2-3=8-3=5\)

\(\text{f(1)=4.1-3=4-3=1}\)

\(\text{f(2)=4.2-3=8-3=5}\)

a: f(2)=9

f(-1/4)=0

\(a,f\left(2\right)=8+1=9\\ f\left(-\dfrac{1}{4}\right)=4\left(-\dfrac{1}{4}\right)+1=-1+1=0\\ b,f\left(x\right)=13\Leftrightarrow4x+1=13\Leftrightarrow x=3\)

a) Ta có: \(y=f\left(x\right)=4x^2-5\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(3\right)=4.3^2-5=31\\f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-5=-4\end{matrix}\right.\)

b) Ta có: \(f\left(x\right)=-1\)

\(\Rightarrow4x^2-5=-1\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1\right\}\) thì \(f\left(x\right)=-1\)

c) \(\forall x\in R,f\left(x\right)=f\left(-x\right)\Leftrightarrow f\left(-x\right)=4.\left(-x\right)^2-5=4x^2-5=f\left(x\right)\)

Vậy \(\forall x\in R\) thì \(f\left(x\right)=f\left(-x\right)\)

\(a.f\left(3\right)=4.3^2-5=31.\\ f\left(\dfrac{-1}{2}\right)=4.\left(\dfrac{-1}{2}\right)^2-5=-4.\)

\(b.f\left(x\right)=-1.\Rightarrow4x^2-5=-1.\\ \Leftrightarrow4x^2=4.\Leftrightarrow x^2=1.\\ \Leftrightarrow x=\pm1.\)

\(c.f\left(x\right)=f\left(-x\right).\\ \Rightarrow4x^2-5=4\left(-x\right)^2-5.\\ \Leftrightarrow4x^2-5=4x^2-5.\)

\(\Leftrightarrow0x=0\) (luôn đúng).

Vậy với mọi x ∈ R thì f (x)= f (-x).

a) Do \( y=f(x)=4x² - 5 \) nên : 

\(+) f(3) = 4 . 3^2 - 5 = 4 . 9 - 5 = 36 - 5 = 31 \) 

\(+) f(\dfrac{1}{2}) = 4 . (\dfrac{1}{2})^2 - 5 = 4 . \dfrac{1}{4} - 5 = 1 - 5 = -4 \) 

Vậy : \(f(3) = 31 ; f(\dfrac{1}{2}) = -4 \) 

b) Do \(f(x) = -1 \) 

Mà \(f(x) = 4x^2 - 5 \) 

\(=> \) \(4x^2 - 5 = -1 \)

\(=> 4x^2 = -1 + 5 \) 

\(=> 4x^2 = 4 \) 

\(=> x^2 = 1 \) \(= 1^2 = ( -1)^2 \) 

\(=> x \) ∈ { -1 ; 1 }

Vậy với \(f(x) = -1 \) thì x ∈ { -1 ; 1 } 

c) Ta có : Do \(x^2 = ( -x )^2 \) 

\(=> \) \(4x^2 = 4(-x)^2 \) 

\(=> 4x^2 - 5 = 4( -x )^2 - 5 \) 

\(=> f(x) = f(-x) \)

Vậy với mọi x ∈ R thì \(f(x) = f(-x)\)