Identify the type of function.

Here are the domains of definition for each function: 1) f(x)=-5 Step 1: Identify the type of function. This is a constant function, which is a type of polynomial. Polynomial functions are defined for all real numbers. The domain of f(x)=-5 is R. 2) f(x)=3x-4 Step 1: Identify the type of function. This is a linear function, which is a type of polynomial. Polynomial functions are defined for all real numbers. The domain of f(x)=3x-4 is R. 3) f(x)=-3x^4-7x^2+2x+5 Step 1: Identify the type of function. This is a polynomial function. Polynomial functions are defined for all real numbers. The domain of f(x)=-3x^4-7x^2+2x+5 is R. 4) f(x)=(2x+1)/((x-1)^2-4) Step 1: Identify restrictions. For a rational function, the denominator cannot be equal to zero. Set the denominator to zero and solve for x: (x-1)^2 - 4 = 0 Step 2: Solve for x. (x-1)^2 = 4 Take the square root of both sides: x-1 = ±sqrt(4) x-1 = ± 2 This gives two possible values for x: x-1 = 2 x = 3 x-1 = -2 x = -1 So, x 3 and x -1. The domain of f(x)=(2x+1)/((x-1)^2-4) is R \-1, 3\. 5) f(x)=sqrt(-4x(x+2)) Step 1: Identify restrictions. For a square root function, the expression under the square root must be greater than or equal to zero. -4x(x+2) 0 Step 2: Find the critical points. Set -4x(x+2) = 0. This gives x=0 or x+2=0 x=-2. The critical points are x=-2 and x=0. Step 3: Test intervals. We test the intervals (-, -2], [-2, 0], and [0, ). For x < -2 (e.g., x=-3): -4(-3)(-3+2) = 12(-1) = -12 < 0. For -2 x 0 (e.g., x=-1): -4(-1)(-1+2) = 4(1) = 4 0. For x > 0 (e.g., x=1): -4(1)(1+2) = -4(3) = -12 < 0. The inequality -4x(x+2) 0 is satisfied when -2 x 0. The domain of f(x)=sqrt(-4x(x+2)) is [-2, 0]. 6) f(x)=(-6x)/((x+3)(4-x)) Step 1: Identify restrictions. For a rational function, the denominator cannot be equal to zero. Set the denominator to zero and solve for x: (x+3)(4-x) = 0 This gives two possible values for x: x+3 = 0 x = -3 4-x = 0 x = 4 So, x -3 and x 4. The domain of f(x)=(-6x)/((x+3)(4-x)) is R \-3, 4\. 7) f(x)=(x+3)/(x^2-7x) Step 1: Identify restrictions. For a rational function, the denominator cannot be equal to zero. Set the denominator to zero and solve for x: x^2 - 7x = 0 Step 2: Factor the denominator. x(x-7) = 0 This gives two possible values for x: x = 0 x-7 = 0 x = 7 So, x 0 and x 7. The domain of f(x)=(x+3)/(x^2-7x) is R \0, 7\. 8) f(x)=sqrt((2x-5)/(3-x)) Step 1: Identify restrictions. For a square root of a rational function, two conditions must be met: 1. The expression under the square root must be non-negative: (2x-5)/(3-x) 0. 2. The denominator cannot be zero: 3-x 0. Step 2: Solve the inequality (2x-5)/(3-x) 0. Find the critical points by setting the numerator and denominator to zero: 2x-5 = 0 2x = 5 x = (5)/(2) = 2.5 3-x = 0 x = 3 Step 3: Test intervals. We test the intervals (-, 2.5], [2.5, 3), and (3, ). Note that x=3 is excluded due to the denominator. For x < 2.5 (e.g., x=0): (2(0)-5)/(3-0) = (-5)/(3) < 0. For 2.5 x < 3 (e.g., x=2.8): (2(2.8)-5)/(3-2.8) = (5.6-5)/(0.2) = (0.6)/(0.2) = 3 0. For x > 3 (e.g., x=4): (2(4)-5)/(3-4) = (8-5)/(-1) = (3)/(-1) = -3 < 0. The inequality (2x-5)/(3-x) 0 is satisfied when 2.5 x < 3. The domain of f(x)=sqrt((2x-5)/(3-x)) is [(5)/(2), 3). 9) f(x)=(x+3)/(x^2+1) Step 1: Identify restrictions. For a rational function, the denominator cannot be equal to zero. Set the denominator to zero and solve for x: x^2 + 1 = 0 Step 2: Solve for x. x^2 = -1 There are no real solutions for x^2 = -1, as the square of any real number is non-negative. Therefore, the denominator x^2+1 is never zero for any real x. The domain of f(x)=(x+3)/(x^2+1) is R. 10) f(x)=(x+2)/(sqrt((6-2x)(x+9))) Step 1: Identify restrictions. For a square root in the denominator, the expression under the square root must be strictly greater than zero (cannot be zero because it's in the denominator). (6-2x)(x+9) > 0 Step 2: Find the critical points. Set (6-2x)(x+9) = 0. This gives: 6-2x = 0 2x = 6 x = 3 x+9 = 0 x = -9 The critical points are x=-9 and x=3. Step 3: Test intervals. We test the intervals (-, -9), (-9, 3), and (3, ). For x < -9 (e.g., x=-10): (6-2(-10))(-10+9) = (6+20)(-1) = 26(-1) = -26 < 0. For -9 < x < 3 (e.g., x=0): (6-2(0))(0+9) = (6)(9) = 54 > 0. For x > 3 (e.g., x=4): (6-2(4))(4+9) = (6-8)(13) = (-2)(13) = -26 < 0. The inequality (6-2x)(x+9) > 0 is satisfied when -9 < x < 3. The domain of f(x)=(x+2)/(sqrt((6-2x)(x+9))) is (-9, 3). That's 2 down. 3 left today — send the next one.