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a: \(f\left(-1\right)=3-7=-4\)
\(f\left(\dfrac{1}{5}\right)=\dfrac{3}{25}-7=\dfrac{-172}{25}\)
b: f(x)=-20/3
\(\Leftrightarrow3x^2-7=-\dfrac{20}{3}\)
\(\Leftrightarrow3x^2=\dfrac{1}{3}\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
=>x=1/3 hoặc x=-1/3
\(a.\)
Theo đề , ta có : \(y=f\left(x\right)=4x^2-5\)
\(\Rightarrow\)
\(f\left(3\right)=4.\left(3\right)^2-5=31\)
\(f\left(-\frac{1}{2}\right)=4.\left(-\frac{1}{2}\right)^2-5=-4\)
\(b.\)
Ta có : \(f\left(x\right)=-1\)
\(\Rightarrow4x^2-5=-1\)
\(\Rightarrow4x^2=-1+5=4\)
\(\Rightarrow x^2=4:4=1\)
\(\Rightarrow x=\sqrt{1}=1\)
\(c.\)
Ta có :
\(f\left(x\right)=4x^2-5\)
\(\Rightarrow f\left(x\right)=4.\left(x\right)^2-5\) \(\left(1\right)\)
\(f\left(-x\right)=4.\left(-x\right)^2-5=4.\left(x\right)^2-5\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow f\left(x\right)=f\left(-x\right)\)
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.
Ta có y=f(x)=2x2+2016
\(f\left(\sqrt{2}\right)=2.\left(\sqrt{2}\right)^2+2016=2.2+2016=4+2016=2020\)
\(f\left(\frac{\sqrt{2}}{2}\right)=2.\left(\frac{\sqrt{2}}{2}\right)^2+2016=2.1+2016=2+2016=2018\)
\(f\left(-\frac{1}{2}\right)=2.\left(-\frac{1}{2}\right)^2+2016=2\cdot\frac{1}{4}+2016=\frac{1}{2}+2016=0.5+2016=2016,5\)
vậy.....
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:\(f\left(x\right)=0\Rightarrow|3x-1|=0\Rightarrow3x-1=0\)
\(3x=0+1=1\)
\(x=1:3=\dfrac{1}{3}\)
\(f\left(x\right)=1\Rightarrow|3x-1|=1\Rightarrow3x-1=\pm1\)
*Với \(3x-1=1\Rightarrow3x=1+1=2\)
\(x=2:3=\dfrac{2}{3}\)
*Với \(3x-1=-1\Rightarrow3x=-1+1=0\)
\(x=0:3=0\)
\(f\left(x\right)=\dfrac{1}{2}\Rightarrow|3x-1|=\dfrac{1}{2}\Rightarrow3x-1=\pm\dfrac{1}{2}\)
*Với \(3x-1=\dfrac{1}{2}\Rightarrow3x=\dfrac{1}{2}+1=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:3=\dfrac{3}{2}.\dfrac{1}{3}=\dfrac{1}{2}\)
*Với \(3x-1=-\dfrac{1}{2}\Rightarrow3x=-\dfrac{1}{2}+1=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}:3=\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}\)
\(f\left(x\right)=-\dfrac{2010}{2011}\Rightarrow|3x-1|=-\dfrac{2010}{2011}\Rightarrow x\in\varnothing\)
a) \(f\left(3\right)=4\times3^2-5=31\)
\(f\left(-\frac{1}{2}\right)=4\times\left(-\frac{1}{2}\right)^2-5=-4\)
b) để f(x)=-1
<=>\(4x^2-5=-1\)
<=>\(4x^2=4\)
<=>\(x^2=1\)
<=>\(x=\orbr{\begin{cases}1\\-1\end{cases}}\)
Cho hàm số y = f(x) = 4x^2 +4y=f(x)=4x2+4. Tính f(-2)f(−2) ; f(2)f(2) ; f(4)f(4).
Đáp số:
f(-2) =f(−2)=
f(2) =f(2)=
f(4) =f(4)=