Answer:
10)[tex] (f \cdot g)(x) = -6x^3 + x^2 + 24x - 4 [/tex]
11) [tex] \left(\dfrac{h}{f}\right) = \dfrac{2x^2 - 9x + 2}{1 - 6x} [/tex] restriction: [tex] x = \dfrac{1}{6} [/tex]
12) [tex] (g \circ f)(x) = 36x^2 - 12x - 3 [/tex]
Step-by-step explanation:
Problem 10: [tex] (f \cdot g)(x) [/tex]
To find [tex] (f \cdot g)(x) [/tex], we need to multiply [tex] f(x) [/tex] and [tex] g(x) [/tex].
Given:
We compute:
[tex] (f \cdot g)(x) = f(x) \cdot g(x) [/tex]
[tex] = (1 - 6x)(x^2 - 4) [/tex]
[tex] = x^2 - 4 - 6x^3 + 24x [/tex]
[tex] = -6x^3 + x^2 + 24x - 4 [/tex]
So, [tex] (f \cdot g)(x) = -6x^3 + x^2 + 24x - 4 [/tex].
There are no domain restrictions for this function.
Problem 11: [tex] \left(\dfrac{h}{f}\right)(x) [/tex]
To find [tex] \left(\dfrac{h}{f}\right)(x) [/tex], we need to divide [tex] h(x) [/tex] by [tex] f(x) [/tex].
Given:
We compute:
[tex] \left(\dfrac{h}{f}\right)(x) = \dfrac{h(x)}{f(x)} [/tex]
[tex] = \dfrac{2x^2 - 9x + 2}{1 - 6x} [/tex]
To find the domain, we need to make sure that the denominator [tex] 1 - 6x [/tex] is not equal to zero:
[tex] 1 - 6x \neq 0 [/tex]
[tex] 6x \neq 1 [/tex]
[tex] x \neq \dfrac{1}{6} [/tex]
So, the domain of [tex] \left(\dfrac{h}{f}\right)(x) [/tex] is all real numbers except [tex] x = \dfrac{1}{6} [/tex].
Problem 12: [tex] (g \circ f)(x) [/tex]
To find [tex] (g \circ f)(x) [/tex], we need to perform the composition [tex] g(f(x)) [/tex].
Given:
We compute:
[tex] (g \circ f)(x) = g(f(x)) [/tex]
[tex] = g(1 - 6x) [/tex]
[tex] = (1 - 6x)^2 - 4 [/tex]
[tex] = (1 - 12x + 36x^2) - 4 [/tex]
[tex] = 36x^2 - 12x - 3 [/tex]
So, [tex] (g \circ f)(x) = 36x^2 - 12x - 3 [/tex].
There are no domain restrictions for this function.
