Given f(x)=-3x^2-7x+19 and g(x)=2x^2-4x-5 find the following:a. (f-g)(-2) b. (fg)(-2) c. (g°f)(-2) d. g-1(-2) e. f(g(f(2)))Supposedly there are two different answers for question "d". Details!...

Given f(x)=-3x^2-7x+19 and g(x)=2x^2-4x-5 find the following:a. (f-g)(-2) b. (fg)(-2) c. (g°f)(-2) d. g-1(-2) e. f(g(f(2)))

Supposedly there are two different answers for question "d". Details! Details! I need all the steps, that way I might be better able to understand it all! Thanks!

Asked on by kakes

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

a) (f-g)(x) = (-3x^2-2x^2)+(-7x+4x)+(19+5)

(f-g)(x) = -5x^2-3x+24

 

b) For calculating (fg)(-2), we'll calculate first the product (fg)(x).

(fg)(x)= (-3x^2-7x+19 )*(2x^2-4x-5)

We'll multiply each term from the first bracket with each term from the second one and the resul will be:

(fg)(x)= -6x^4+12x^3+15x^2-14x^3+28x^2+35x+38x^2-76x-95

We'll group the terms which have the same unknown raised to the same degree;

(fg)(x)= -6x^4+x^3(12-14)+x^2*(15+28+38)+x*(35-76)-95

(fg)(x)= -6x^4 - 2x^3 + 52x^2 -41x -95

Now, we'll calculate fg(-2), by substituting x with the value (-2):

(fg)(-2)= -6(-2)^4 - 2(-2)^3 + 52(-2)^2 -41(-2) -95

(fg)(-2)= -96+16+186-95

(fg)(-2)= -11

 

c) To calculate (g°f)(-2), first we have to calculate (g°f)(x).

(g°f)(x) = g(f(x)) = 2*[f(x)]^2-4*f(x)-5

g(f(x))=2(9x^4+49+x^2+361-42x^3-57x^2-133x)+12x^2+28x-81

g(f(2))= 2(144+196+361-336-228-266)+48+56-81

g(f(2))= -235

d) (g-1)(x) = 2x^2-4x-5-1

(g-1)(x) = 2x^2-4x-6

We'll calculate (g-1)(x) as we have calculated f-g, but this time having a constant function =1, instead a mono variable function.

(g-1)(-2) = 2(-2)^2-4(-2)-6

(g-1)(-2) = 8+8-6

(g-1)(-2) = 16-6

(g-1)(-2) = 10

e. For calculating f(g(f(2))), we'll use the result from the point c), where we've calculated already g(f(2))= -235.

So, we'll have to calculate f(-235).

f(-235) = -3(-235)^2 - 7(-235) + 19

f(-235) = -165675 + 1645 + 19

f(-235) = -164011

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

a)

(f-g)(x) = {f(x)-g(x)} = -3x^2-7x+19-(2x^2-4x-5)

=-5x^2-7x+4x+19+5 = -5x^2-3x+24.

(f-g)(-2)=-5(-2)^2-3(-2)+24 = -5*4+6+24 = -2

b) fg(-2) = (-3x^2-7x+19)(2x^2-4x-5) when x =-2

=[(-3(-2)^2-7(-2)+19][2(-2)^2-4(-2)-5]

=[-12+14+19][8+8-5] = 21*11 = 231

c)

gof(-2) = g(f(-2)) = g(2(-2)^2-4(-2)-5) = g(11) = -3(11)^2-7(11)+19 =  =-363+77+19 =-267

d) g inverse (2) = k then -2 = g(k) = 2k^2-4k-5. Or

2k^2-4k-3 = 0. Or k = [4+or-sqrt(4^2+4*2*3)]/4 = 1+or-sqrt(2*5). Or k = 1+srt(5/2) Or 1-sqrt(5/2).

e) f(g(f(2) = f(g(-3*2^2-7*2+19)

=f(g(-7)) = f{2(-7)^2-4(-7)-5} = f(121) = -3(121)^2-7(121)+19= -15348

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