#### To determine

**To find:** The derivative of the function f(x)=x2−12x−3 and state the domain of the function and its derivative.

#### Answer

The derivative of the function f(x) is 2x2−6x+2(2x−3)2_.

The domain of the function f(x) is (−∞,32)∪(32,∞).

The domain of the derivative f′(x) is (−∞,32)∪(32,∞).

#### Explanation

**Formula used:**

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

f′(x)=limh→0f(x+h)−f(x)h (1)

**Calculation:**

Obtain the derivative of the function f(x)

Use the equation (1) to compute f′(x).

f′(x)=limh→0f(x+h)−f(x)h=limh→0((x+h)2−12(x+h)−3)−(x2−12x−3)h=limh→0((x+h)2−1)(2x−3)−(x2−1)(2(x+h)−3)(2(x+h)−3)(2x−3)h=limh→0(x2+h2+2xh−1)(2x−3)−(x2−1)(2x+2h−3)h(2x+2h−3)(2x−3)

Expand the numerator and simplifying further as follows,

f′(x)=limh→0((2x(x2+h2+2xh−1)−3(x2+h2+2xh−1)−x2(2x+2h−3)+1(2x+2h−3))h(2x+2h−3)(2x−3))=limh→0((2x3+2xh2+4x2h−2x−3x2−3h2−6xh+3−2x3−2x2h+3x2+2x+2h−3)h(2x+2h−3)(2x−3))=limh→0(2xh2+2x2h−3h2−6xh+2hh(2x+2h−3)(2x−3))=limh→0(h(2xh+2x2−3h−6x+2)h(2x+2h−3)(2x−3))

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

f′(x)=limh→0((2xh+2x2−3h−6x+2)(2x+2h−3)(2x−3))=((2x(0)+2x2−3(0)−6x+2)(2x+2(0)−3)(2x−3))=(2x2−6x+2)(2x−3)(2x−3)=(2x2−6x+2)(2x−3)2

Thus, the derivative of the function f(x) is 2x2−6x+2(2x−3)2_.

The function is defined for every real numbers except x=32.

Therefore, the domain of the function is (−∞,32)∪(32,∞).

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

Since the derivative f′(x)=2x2−6x+2(2x−3)2 exists for all real numbers except x=32.

Therefore, the domain of f′(x) is (−∞,32)∪(32,∞).