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Ta có hàm số sau :
\(f\left(1\right)=3.1^2-1=2\)
\(f\left(\frac{-2}{3}\right)=3.\frac{-2}{3}-1=-2-1=-3\)
Vậy hàm số f(1) = 2
Hàm số :\(f\left(\frac{-2}{3}\right)=-3\)
Ta có y = f(x) = 3x2 + 1. Do đó
f(\(\dfrac{1}{2}\)) = 3.\(\left(\dfrac{1}{2}\right)^2\) + 1 = \(\dfrac{3}{4}\)+ 1 = \(\dfrac{7}{4}\)
f(1) = 3.12 + 1 = 3.1 + 1 = 3 + 1 = 4
f(3) = 3.32 + 1 = 3.9 + 1 = 27 + 1 = 28.
\(f\left(1\right)=-\dfrac{3}{2}.1=-\dfrac{3}{2}\)
\(f\left(-1\right)=-\dfrac{3}{2}.\left(-1\right)=\dfrac{3}{2}\)
\(f\left(2\right)=-\dfrac{3}{2}.2=-3\)
\(f\left(-2\right)=-\dfrac{3}{2}.\left(-2\right)=3\)
\(f\left(\dfrac{1}{2}\right)=-\dfrac{3}{2}.\dfrac{1}{2}=\dfrac{-3}{4}\)
\(f\left(-\dfrac{1}{2}\right)=-\dfrac{3}{2}.\left(-\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(f\left(a\right)< f\left(-a\right)\)
f (1) = 2 . 12 - 5 = -3
f (-2) = 2 . (-2)2 - 5 = 3
f (0) = 2 . 02 - 5 = -5
f (2) = 2 . 22 - 5 = 3
Có: \(f\left(x\right)=2x^2-5\)
\(\Rightarrow f\left(1\right)=2.1^2-5=-3\)
\(f\left(-2\right)=2.\left(-2\right)^2-5=3\)
\(f\left(0\right)=2.0^2-5=-5\)
\(f\left(2\right)=2.2^2-5=3\)
\(•f\left(0\right)=2.0^2-3.0=0\\ •f\left(2\right)=2.2^2-3.2=2\\ •f\left(\dfrac{1}{2}\right)=2.\left(\dfrac{1}{2}\right)^2.3.\dfrac{1}{2}=-1\\ •f\left(\dfrac{-2}{3}\right)=2.\left(-\dfrac{2}{3}\right)^2-3.\left(-\dfrac{2}{3}\right)=\dfrac{26}{9}\\ •f\left(3\right)=2.3^2-3.3=9\)
a) Thay f(0);f(\(-\frac{1}{2}\)) vào f(x)=2-x2 ta được:
\(f\left(0\right)=2-0^2=2\)
\(f\left(-\frac{1}{2}\right)=2-\left(-\frac{1}{2}\right)^2=\frac{7}{4}\)
b) y = f(x) = 2-x2
Ta có f(x-1) = 2- (x-1)2
f(1-x) = 2 - (1-x)2 = 2 - (x-1)2
nên f(x-1) = f(1-x)
a)Ta có : \(y=f\left(x\right)=2x^2-\dfrac{1}{4}\)
\(\Rightarrow y=f\left(0\right)=2.0^2-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(f\left(\dfrac{-1}{2}\right)=2.\left(\dfrac{-1}{2}\right)^2-\dfrac{1}{4}=2.\dfrac{1}{4}-\dfrac{1}{4}=\dfrac{1}{4}\)
b)Ta có : \(\dfrac{27}{5}=2x^2-\dfrac{1}{4}\)
\(\Rightarrow2x^2=\dfrac{27}{5}+\dfrac{1}{4}=\dfrac{113}{20}\)
\(\Rightarrow x^2=\dfrac{113}{40}\)
\(\Rightarrow x=\sqrt{\dfrac{113}{40}}\)
Sửa đề: \(y=f\left(x\right)=2x^2-\dfrac{1}{4}\)
a)\(\Rightarrow\left\{{}\begin{matrix}f\left(0\right)=2.0^2-\dfrac{1}{4}=0-\dfrac{1}{4}=-\dfrac{1}{4}\\f\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^2-\dfrac{1}{4}=\dfrac{1}{4}-\dfrac{1}{4}=0\end{matrix}\right.\)
b) \(y=\dfrac{27}{5}\Leftrightarrow2x^2-\dfrac{1}{4}=\dfrac{27}{5}\)
\(\Leftrightarrow2x^2=\dfrac{27}{5}+\dfrac{1}{4}=\dfrac{113}{20}\Leftrightarrow x^2=\dfrac{113}{40}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{113}{40}}\\x=-\sqrt{\dfrac{113}{40}}\end{matrix}\right.\)
\(f\left(x\right)=2x+\dfrac{1}{2}\)
a) \(f\left(0\right)=2.0+\dfrac{1}{2}=0+\dfrac{1}{2}=\dfrac{1}{2}\)
b) \(f\left(\dfrac{1}{2}\right)=2.\dfrac{1}{2}+\dfrac{1}{2}=1+\dfrac{1}{2}=\dfrac{3}{2}\)
c) \(f\left(-2\right)=2.\left(-2\right)+\dfrac{1}{2}=-4+\dfrac{1}{2}=\dfrac{-7}{2}\)
f(x)=2x+12f(x)=2x+12
a) f(0)=2.0+12=0+12=12f(0)=2.0+12=0+12=12
b) f(12)=2.12+12=1+12=32f(12)=2.12+12=1+12=32
c) f(−2)=2.(−2)+12=−4+12=−72