Question

[One Piece]

Hàm số:

1. Cho f(x)=\(\frac{2}{3}-\frac{1}{2}\left|x-1\right|\)

Tính f(0); f(-1); f(1); f(\(\frac{3}{4}\))

2. Cho f(x)=ax+b. Tìm a,b biết

a.f(0)=3; f(1)=-5

b.f(1)=5; f(-1)=2

Answer 1

[Trần Nguyễn Bảo Quyên]

\(.1.\)

Ta có : \(f\left(x\right)=\frac{2}{3}-\frac{1}{2}\left|x-1\right|\)

Thay

+ \(f\left(0\right)=\frac{2}{3}-\frac{1}{2}\left|0-1\right|\)

\(\Rightarrow f\left(0\right)=\frac{2}{3}-\frac{1}{2}\)

\(\Rightarrow f\left(0\right)=\frac{4}{6}-\frac{3}{6}\)

\(\Rightarrow f\left(0\right)=\frac{1}{6}\)

+ \(f\left(-1\right)=\frac{2}{3}-\frac{1}{2}\left|-1-1\right|\)

\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{1}{2}.2\)

\(\Rightarrow f\left(-1\right)=\frac{2}{3}-1\)

\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{3}{3}\)

\(\Rightarrow f\left(-1\right)=-\frac{1}{3}\)

+ \(f\left(1\right)=\frac{2}{3}-\frac{1}{2}\left|1-1\right|\)

\(\Rightarrow f\left(1\right)=\frac{2}{3}-\frac{1}{2}.0\)

\(\Rightarrow f\left(1\right)=\frac{2}{3}-0\)

\(\Rightarrow f\left(1\right)=\frac{2}{3}\)

+ \(f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}\left|\frac{3}{4}-1\right|\)

\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}.\frac{1}{4}\)

\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{8}\)

\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{16}{24}-\frac{3}{24}=\frac{13}{24}\)