Question

question 3 of 15
for the real-valued functions $f(x)=x^{2}+2$ and $g(x)=x^{2}+4$, find the composition $f \circ g$. also, specify its domain using interval notation.
$(f \circ g)(x)=$
domain of $f \circ g$:

for the real-valued functions $f(x)=x^{2}+2$ and $g(x)=x^{2}+4$, find the composition $f \circ g$. also, specify its domain using interval notation.
$f \circ g\\ x =$
domain of $f \circ g$:

question 4 of 15
find the difference quotient $\frac{f(x+h)-f(x)}{h}$, where $h \
eq 0$, for the function below.
$f(x)=-2x^{2}-3x+6$
simplify your answer as much as possible.

question 5 of 15
find the difference quotient $\frac{f(x+h)-f(x)}{h}$, where $h \
eq 0$, for the function below.
$f(x)=3x^{2}+8$
simplify your answer as much as possible.

Explanation:

Question 3

Step1: Define composition

$(f \circ g)(x) = f(g(x))$

Step2: Substitute $g(x)$ into $f$

$f(g(x)) = (g(x))^2 + 2 = (x^2 + 4)^2 + 2$

Step3: Expand and simplify

$(x^2 + 4)^2 + 2 = x^4 + 8x^2 + 16 + 2 = x^4 + 8x^2 + 18$

Step4: Determine domain

Both $f(x)$ and $g(x)$ are polynomials, so all real numbers are allowed.

Step1: Compute $f(x+h)$

$f(x+h) = -2(x+h)^2 - 3(x+h) + 6 = -2(x^2 + 2xh + h^2) - 3x - 3h + 6 = -2x^2 -4xh -2h^2 -3x -3h +6$

Step2: Calculate $f(x+h)-f(x)$

$f(x+h)-f(x) = (-2x^2 -4xh -2h^2 -3x -3h +6) - (-2x^2 -3x +6) = -4xh -2h^2 -3h$

Step3: Divide by $h$

$\frac{f(x+h)-f(x)}{h} = \frac{h(-4x -2h -3)}{h} = -4x -2h -3$

Step1: Compute $f(x+h)$

$f(x+h) = 3(x+h)^2 + 8 = 3(x^2 + 2xh + h^2) + 8 = 3x^2 + 6xh + 3h^2 + 8$

Step2: Calculate $f(x+h)-f(x)$

$f(x+h)-f(x) = (3x^2 + 6xh + 3h^2 + 8) - (3x^2 + 8) = 6xh + 3h^2$

Step3: Divide by $h$

$\frac{f(x+h)-f(x)}{h} = \frac{h(6x + 3h)}{h} = 6x + 3h$

Answer:

$(f \circ g)(x) = x^4 + 8x^2 + 18$
Domain of $f \circ g$: $(-\infty, \infty)$

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Question 4