Verified answer

(f - g)(x)= x²- 3 - (20 - x)

= x²- 3 - 20 + x

= x²- 23 + x

for x=4

(f - g)(4)= 4^2 - 23 + 4

= 16 - 23 + 4

= - 3 Answer

If f(x) = x²-3 and g(x) = 20 - x then

f(x) - g(x) = (x²-3)-(20-x) = x²-3-20+x (expanding the bracket and multiplying everything in the bracket by -1

You also know that f(x) - g(x) = (f-g)(x)

So, (f-g)(x) =x²-3-20+x

Therefore (f-g)(4) = 4²-3-20+4 = 16-3-20+4 = -3

the least complicated thank you to remedy the 1st would be to do away with the fractions. to try this, we are going to would desire to multiply by capacity of the Least effortless assorted of the denominators of all the fractions. Our denominators are 3, 4, and 3, so the LCM is 12. in case you multiply the entire equation by capacity of 12, all the fractions will disappear. [6x - 4/3 = 3/4x + a million/3](12) 72x - 40 8/3 = 36/4x + 12/3 72x - sixteen = 9x + 4 -9x -9x 63x - sixteen = 4 +sixteen +sixteen 63x = 20 x = 20/sixty 3 the only way that's incorrect is in case your 3/4x is actual 3/(4x), which capacity the x is on the backside. i presumed you meant (3/4)x, with the x the two on good or top next to the fraction. no remember if it is meant to be on the backside, then you definately would desire to multiply by capacity of 12x to get all the fractions to cancel, and you gets this: [6x - 4/3 = 3/4x + a million/3](12x) 72x^2 - 16x = 9 + 4x 72x^2 - 20x - 9 = 0 This gets enormously nasty, so i'm guessing the 1st one is nice. 5(x + a million) = 7x + 3 distribute the 5 to the x and the a million 5x + 5 = 7x + 3 -5x -5x 5 = 2x + 3 -3 -3 2 = 2x a million = x

-3

f(x) = x²-3 and g(x) =20-x

(f-g)(x)= x²-3 -(20-x )

= x^3+x-23

(f-g)(x)= x^3+x-23

(f-g)(4)= 4^3+4-23

= 64+4-23

= 68-23

= 45

(f-g)(4) = -84

For additional information, please see also:

http://en.wikipedia.org/wiki/Function(mathematics)