Question 4Let f be the function defined by the formula,f(x) = 1/x+1/x − 10.a) Determine the largest possible domain D of f.b) Is f injective on D?[8,5]Question 5Compute the f ◦ g and its range of the functions f and g below,f(x) = (x^2 + 5x − 6)(x^2 + 5)/|2x + 3|, and g(x) = √x + 4[12]Question 6Determine the largest domain, intersection with axes, and sign of ff(x) = log2(2 −2/x − 3)[16]

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4.

a)

$D=\{x\in (-\infty ,0)\vee (0,10)\vee (10,\infty )\}$

b)

an injective function is a function f that maps distinct elements to distinct elements; that is,

f(x1)=f(x2)f(x_1)=f(x_2)f(x1​)=f(x2​) implies x1=x2x_1=x_2x1​=x2​

So, f(x) is injective.

5.

f∘g=(6x−2)(x+9)∣2x+4+3∣fcirc g=frac{(6x-2)(x+9)}{|2sqrt{x+4}+3|}f∘g=∣2x+4​+3∣(6x−2)(x+9)​

Range of f(x): $(-\infty ,\infty )$

Range of g(x): $[0,\infty )$

for $f\circ g$ :

$x\in [-4,\infty )$

f∘g(−4)=−22⋅53=−1103fcirc g(-4)=frac{-22cdot 5}{3}=-frac{110}{3}f∘g(−4)=3−22⋅5​=−3110​

Range of $f\circ g$ : $[-110/3,\infty )$

6.

for f(x):

2−2x−3>0  ⟹  2x−8x−3>02-frac{2}{x-3}>0implies frac{2x-8}{x-3}>02−x−32​>0⟹x−32x−8​>0

domain: $x\in (-\infty ,3)\vee (4,\infty )$

for x-intersection:

2−2x−3=1  ⟹  x=52-frac{2}{x-3}=1implies x=52−x−32​=1⟹x=5

x-intersection: $(5,0)$

for y-intersection:

$log$