Question

Dada la función f(x) = 3x2 + 2x encuentre la imagen de

1)F(5)              2) f( -3)               3) f(0)                   4) f( 2/5)              5) f( a+1)              

   6)  f(2a-3)   7) f( 1 +h )- f(1)    8) 2f(3+h) –f(6)​

Answer 1

Respuesta:

Para encontrar la imagen de la función simplemente reemplazamos a x por el valor que está entre paréntesis

$\bold{f(x)=3x^2+2x}\\\\\\1)f(5)=3(5)^2+2(5)\quad\to f(5)=3(25)+10\quad\to f(5)=75+10\quad\to f(5)=85\\\\\\2)f(-3)=3(-3)^2+2(-3)\to f(-3)=3(9)-6\to f(-3)=27-6\to f(-3)=21\\\\\\3)f(0)=3(0)^2+2(0)\quad\to f(0)=3(0)+0\quad\to f(0)=0+0\quad\to f(0)=0\\\\\\4)f(\frac{2}{5})=3(\frac{2}{5})^2+2(\frac{2}{5})\quad\to f(\frac{2}{5} )=3(\frac{4}{25})+\frac{4}{5}\quad\to f(\frac{2}{5})=\frac{12}{25}+\frac{4}{5}\quad\to f(\frac{2}{5})=\frac{32}{25}$

$5)f(a+1)=3(a+1)^2+2(a+1)\to f(5)=3(a^2+2a+1)+2(a+1)\\\\\to f(a+1)=3a^2+6a+3+2a+2\quad\to f(a+1)=3a^2+8a+6\\\\\\\\6)f(2a-3)=3(2a-3)^2+2(2a-3)\to f(5)=3(4a^2-6a+9)+2(2a-3)\\\\\to f(2a-3)=12a^2-18a+27+4a-6\quad\to f(2a-3)=12a^2-14a+21$

$7)f(1+h)-f(1)=[3(1+h)^2+2(1+h)]-[3(1)^2+2(1)]\\\\f(1+h)-f(1)=[3(1+2h+h^2)+2+2h]-[3+2]\\\\f(1+h)-f(1)=[3+6h+3h^2+2+2h]-[5]\\\\f(1+h)-f(1)=[5+8h+3h^2]-[5]\\\\f(1+h)-f(1)=8h+3h^2$

$8)2f(3+h)-f(6)=2[3(3+h)^2+2(3+h)]-[3(6)^2+2(6)]\\\\2f(3+h)-f(6)=2[3(9+6h+h^2)+6+2h]-[3*36+12]\\\\2f(3+h)-f(6)=2[27+18h+3h^2+6+2h]-[108+12]\\\\2f(3+h)-f(6)=2[33+20h+3h^2]-[120]\\\\2f(3+h)-f(6)=66+40h+6h^2-120\\\\2f(3+h)-f(6)=-54+40h+6h^2$

Espero que te sirva, salu2!!!!