functions f and g are defined by f:x=3x and g:x=x+5.Another function h is such that hgf:x=9x^2+33x+31.find h?

Question

if anyone can help, that will help me a lot, thx in advance....

Answer 1

f(x) = 3x
g(x) = x + 5

h(g(f(x))) = 9x^2 + 33x + 31

First, g(f(x)) = g(3x) = 3x + 5

We don't know what h(x) is, but we know that it has to be a quadratic.

Let h(x) = ax^2 + bx + c. Then

h(g(f(x)) = h(3x + 5) = a[3x + 5]^2 + b[3x + 5] + c

Expand,

= a[9x^2 + 30x + 25] + b[3x + 5] + c
= 9ax^2 + 30ax + 25a + 3xb + 5b + c
= (9a)x^2 + 30ax + 3xb + 25a + 5b + c
= (9a)x^2 + x(30a + 3b) + [25a + 5b + c]

BUT

h(g(f(x))) = 9x^2 + 33x + 31

So we can equate coefficients.

9a = 9
30a + 3b = 33
25a + 5b + c = 31

This yields the solutions a = 1, b = 1, c = 1.

Therefore,

h(x) = x^2 + x + 1

Answer 2

Mathsman is right, but what is obvious to you may not be obvious to me, so, just in case
(don't bother to read on if you already know what to do)
(3x+5)^2
= 9x^2 + 30x +25

9x^2 + 33x + 31
= 9x^2 + 30 x + 25 + 3x + 5 + 1
= (3x + 5)^2 + (3x + 5) + 1
= [gf(x)]^2 + gf(x) + 1

Hence
h(x) = x^2 + x + 1
or, if you like,
h: x becomes x^2 + x + 1

Answer 3

gf: 3x + 5

so:
h(3x + 5) = 9x^2 + 33x + 31

9x^2 + 33x + 31
= 9x^2 + 30x + 3x + 25 + 5 + 1
= (9x^2 + 30x + 25) + (3x + 5) + 1
= (3x + 5)^2 + (3x + 5) + 1

so:
h:x = x^2 + x + 1

Answer 4

gf: x becomes 3x + 5. (Not x = 3x + 5, "becomes" is normally given as an arrow but I can't do that with my keyboard.) Therefore it looks as if h will square gf and perhaps something else. If you work out (3x + 5)^2 you will find that what else you need is fairly obvious.