## Calculus: Early Transcendentals 8th Edition

a/ f(2)=3*4-2+2=12 b/f(-2)=3*$(-2)^{2}$-(-2)+2 = 3*4+2+2 = 16 c/ f(a) = 3$a^{2}$-a+2 d/ f(-a) = 3$(-a)^{2}$-(-a)+2 = 3$a^{2}$+a+2 e/ f(a+1) = 3$(a+1)^{2}$-(a+1)+2 =3$(a+1)^{2}$-a+1 f/ 2f(a) = 2(3$a^{2}$-a+2) = 6$a^{2}$-2a+4 g/ f(2a) = 3$(2a)^{2}$-2a+2 = 12$a^{2}$-2a+2 h/ f($a^{2}$) = 3$($a^{2}$)^{2}$-$a^{2}$+2 = 3$a^{4}$-$a^{2}$+2 i/ $[f(a)]^{2}$ = (3$a^{2}$-a+2)^2 j/ f(a+h) = 3$(a+h)^{2}$-(a+h)+2 = 3$(a+h)^{2}$-a-h+2
f(x) =3$x^{2}$ - x + 2 x is the variable and by replacing x with a number, we can find the value of the function when x equals that number. a/ f(2)=3*4-2+2=12 b/f(-2)=3*$(-2)^{2}$-(-2)+2 = 3*4+2+2 = 16 c/ f(a) = 3$a^{2}$-a+2 d/ f(-a) = 3$(-a)^{2}$-(-a)+2 = 3$a^{2}$+a+2 e/ f(a+1) = 3$(a+1)^{2}$-(a+1)+2 =3$(a+1)^{2}$-a+1 f/ 2f(a) = 2(3$a^{2}$-a+2) = 6$a^{2}$-2a+4 g/ f(2a) = 3$(2a)^{2}$-2a+2 = 12$a^{2}$-2a+2 h/ f($a^{2}$) = 3$(a^{2})^{2}$-$a^{2}$+2 = 3$a^{4}$-$a^{2}$+2 i/ $[f(a)]^{2}$ = (3$a^{2}-a+2)^2$ j/ f(a+h) = 3$(a+h)^{2}$-(a+h)+2 = 3$(a+h)^{2}$-a-h+2