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...

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