Find f(0),f(2),f(-2),f(3),f(√2 ), and f(3t) if :
a) f(x)=3x^2 -2
b) f(x)=√(x+1)
a) f(x)=3x2−2,f(x)=3x^2-2,f(x)=3x2−2,
f(0)=3∗02−2=−2f(0)=3*0^2-2=-2f(0)=3∗02−2=−2
f(2)=3∗22−2=3∗4−2=10f(2)=3*2^2-2=3*4-2=10f(2)=3∗22−2=3∗4−2=10
f(−2)=3∗(−2)2−2=3∗4−2=10f(-2)=3*(-2)^2-2=3*4-2=10f(−2)=3∗(−2)2−2=3∗4−2=10
f(3)=3∗32−2=3∗9−2=25f(3)=3*3^2-2=3*9-2=25f(3)=3∗32−2=3∗9−2=25
f(2)=3∗(2)2−2=3∗2−2=4f(\sqrt{2})=3*(\sqrt{2})^2-2=3*2-2=4f(2)=3∗(2)2−2=3∗2−2=4
f(3t)=3∗(3t)2−2=3∗9t2−2=27t2−2f(3t)=3*(3t)^2-2=3*9t^2-2=27t^2-2f(3t)=3∗(3t)2−2=3∗9t2−2=27t2−2
b) f(x)=x+1,f(x)=\sqrt{x+1},f(x)=x+1,
f(0)=0+1=1f(0)=\sqrt{0+1}=1f(0)=0+1=1
f(2)=2+1=3f(2)=\sqrt{2+1}=\sqrt{3}f(2)=2+1=3
f(−2)=−2+1=−1f(-2)=\sqrt{-2+1}=\sqrt{-1}f(−2)=−2+1=−1 is undetermined in real numbers
f(3)=3+1=4=2f(3)=\sqrt{3+1}=\sqrt{4}=2f(3)=3+1=4=2
f(2)=2+1f(\sqrt{2})=\sqrt{\sqrt{2}+1}f(2)=2+1
f(3t)=3t+1f(3t)=\sqrt{3t+1}f(3t)=3t+1
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