for both problems let: f (x)=x^2 + 4x and g(x)=x+2

first problem:

(f+g)(x) and (f+g)(3)

second problem:

(f-g)(x)and (f-g)(-1)

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- May 30th 2006, 10:42 AM #1

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- May 30th 2006, 11:27 AM #2Originally Posted by
**kbryant05**

1rst problem: $\displaystyle (f+g)(x)=x^2+4x+x+2 = x^2+5x+2$

To calculate (f+g)(x) plug in the value 3 for x:

$\displaystyle (f+g)(3)=3^2+5 \cdot 3+2 = 9+15+2=26$

2nd problem: $\displaystyle (f-g)(x)=x^2+4x-(x+2) = x^2+3x-2$

To calculate (f-g)(-1) plug in the value (-1) for x:

$\displaystyle (f-g)(-1)= (-1)^2+3\cdot (-1)-2=1-3-2=-4$

Greetings

EB

- May 30th 2006, 02:34 PM #3

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