Question

Cho hàm số \(\) \(y=f\left(x\right)=3x^2+1\)

Tính : \(f\left(\dfrac{1}{2}\right);f\left(1\right);f\left(3\right)\)

Answer 1

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.

Answer 2

$y=f\left(x\right)=3x^2+1$

$\mathrm{f\; \backslash (\backslash left(\backslash dfrac\{1\}\{2\}\backslash right)\backslash )=\; 3\; .\; \backslash (\backslash left(\backslash dfrac\{1\}\{2\}\backslash right)^2\backslash )\; +\; 1\; =\; 3\; .\; \backslash (\backslash dfrac\{1\}\{4\}\backslash )\; +\; 1\; =\; \backslash (\backslash dfrac\{3\}\{4\}+1\backslash )\; =\; \backslash (\backslash dfrac\{7\}\{4\}\backslash )}$

f (1) = 3 . 1² + 1= 3 + 1 = 4

f (3) = 3 . 3² + 1 = 3 . 9 + 1 = 28

Answer 3

Ta có : y = f(x) = 3\(x^2+1\)

Khi đó : f\(\left(\dfrac{1}{2}\right)=3.\left(\dfrac{1}{2}\right)^2+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)

\(f\left(1\right)=3.\left(1\right)^2+1=3+1=4\)

\(f\left(3\right)=3.\left(3\right)^2+1=27+1=28\)

Answer 4

Ta có : y = f(x) = 3x^2+1x2+1

f \left(\dfrac{1}{2}\right)(21​)= 3 . \left(\dfrac{1}{2}\right)^2(21​)2 + 1 = 3 . \dfrac{1}{4}41​ + 1 = \dfrac{3}{4}+143​+1 = \dfrac{7}{4}47​

f(1)=3.(1)2+1=3+1=4

f\left(3\right)=3.\left(3\right)^2+1=27+1=28f(3)=3.(3)2+1=27+1=28