#### To determine

**To find:** The derivative of the function g(x)=9−x and state the domain of the function and its derivative.

#### Answer

The derivative of the function g(x) is −129−x_.

The domain of the function g(x) is (−∞,9]

The domain of g′(x) is (−∞,9).

#### Explanation

**Formula used:**

The derivative of a function *f ,* denoted by f′(x), is

f′(x)=limh→0f(x+h)−f(x)h (1)

**Difference of squares formula:** (a2−b2)=(a+b)(a−b)

**Calculation:**

Obtain the derivative of the function g(x).

Use the equation (1) to compute g′(x),

g′(x)=limh→0f(x+h)−f(x)h=limh→0(9−(x+h))−(9−x)h

Multiply both the numerator and the denominator by the conjugate of the numerator,

g′(x)=limh→0(9−(x+h))−(9−x)h×(9−(x+h))+(9−x)(9−(x+h))+(9−x)=limh→0((9−(x+h))−(9−x))((9−(x+h))+(9−x))h((9−(x+h))+(9−x))

Apply the difference of square formula,

g′(x)=limh→0(9−(x+h))2−(9−x)2h((9−(x+h))+(9−x))=limh→09−(x+h)−(9−x)h((9−(x+h))+(9−x))=limh→09−x−h−9+xh((9−(x+h))+(9−x))=limh→0−hh((9−(x+h))+(9−x))

Since the limit *h* approaches zero but not equal to zero, cancel the common term *h* from both the numerator and the denominator,

g′(x)=limh→0−1(9−(x+h))+(9−x)=−19−(x+(0))+9−x=−19−x+9−x=−129−x

Thus, the derivative of the function g(x) is −129−x_.

The function is defined for every real numbers except (9,∞).

Therefore, the domain of the function g(x)=9−x is (−∞,9].

The domain of the derivative g′(x) is {x∈ℝ| g′(x) exists}.

Since the derivative g′(x)=−129−x exists for all real numbers except [9,∞).

Therefore, the domain of g′(x) is (−∞,9).