Question #219078

 If f(x)=3x22x+5\ f(x) = 3x^2 – 2x + 5 , find f[1, 2], f[2, 3] and f[1, 2, 3].


1
Expert's answer
2021-07-20T17:18:50-0400

The function f(x)=3x22x+5f(x)=3x^2-2x+5 is continuous for xRx\in \R as polynomial.



f(x)=6x2f'(x)=6x-2

f(x)=0=>6x2=0=>x=13f'(x)=0=>6x-2=0=>x=\dfrac{1}{3}

If x>13,f(x)>0,f(x)x>\dfrac{1}{3}, f'(x)>0, f(x) is strictly increasing.



f(1)=3(1)22(1)+5=6f(1)=3(1)^2-2(1)+5=6

f(2)=3(2)22(2)+5=13f(2)=3(2)^2-2(2)+5=13

f(3)=3(3)22(3)+5=26f(3)=3(3)^2-2(3)+5=26

f:[1,2][6,26]f:[1,2]\to[6,26]

f:[2,3][13,26]f:[2,3]\to[13,26]

f:{1,2,3}{6,13,26}f:\{1,2,3\}\to\{6, 13,26\}


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