Mac 1105 2015 Test 2 Name You

Mac 1105 2015 1 Test 2 Name You

Determine the intervals over which the function is decreasing, increasing, and constant. Write your answer in interval notation. Evaluate the function for the given values of x. For each function, find specific values, graph the functions, analyze their properties such as intercepts, and determine their domains and ranges. Additionally, perform algebraic operations on functions such as division and subtraction, as well as find the difference quotient, midpoints, distances, and equations of circles in standard form.

Paper For Above instruction

In this analysis, we explore various aspects of multiple functions, including their intervals of increase and decrease, evaluations, graphing, intercepts, and specific point calculations. The first function, a piecewise linear function, facilitates analyzing intervals of monotonicity and specific value calculations. Its form:

f(x) = -3x + 3, for x

f(x) = x2 + 3, for -1 ≤ x

f(x) = 2, for x ≥ 3.

To determine where this function is increasing or decreasing, we evaluate its derivative or analyze the slopes of each piece. For x 2 + 3, which is a parabola opening upward, so it decreases on the interval [-1, 0) and increases on (0, 3). For x ≥ 3, the function is constant at 2, which is neither increasing nor decreasing. For evaluation, f(-1) = (-3)(-1) + 3 = 3 + 3 = 6; and f(4) = 2, since x ≥ 3 and the function value is constant at 2.

Graphing this piecewise function involves plotting the linear segment with a negative slope for x

The second function, f(x) = -5 - x for x

The function g(x) = x2 - 2 is a parabola opening upward, with vertex at (0, -2). Its intercepts are at points where g(x)=0; solving x2 - 2=0 gives x= ±√2. The graph passes through these points, with y-intercept at (0, -2).

Examining symmetry, consider g(x) = -x + 5. To determine if it's even, odd, or neither: evaluate g(-x) = -(-x)+5 = x+5; since g(-x) ≠ g(x) or -g(x), the function is neither even nor odd. Its symmetry doesn't align with these categories.

Using a given graph, specific function values can be estimated or read directly. For instance, if f(x) is graphed, f(-2) can be found by locating x=-2 and reading the y-value. Domain and range depend on the function's definition and the graph's extent.

The domain of a function includes all x-values for which the function is defined. For example, for f(x)=6/(3-x), the domain excludes x=3 where the denominator zeroes out. In interval notation, if the domain is all real numbers except 3, it is (-∞, 3) ∪ (3, ∞).

Functions like f(x)=2x2-9x and g(x)=x2-7x-18 can be combined via division: (f/g)(x) = (2x2-9x)/(x2-7x-18). Simplification involves factoring numerator and denominator; for instance, x2-7x-18 factors into (x-9)(x+2). Determining the domain involves excluding zeros of the denominator.

To find (f - g)(-4), with f(x)=4x2-3 and g(x)=x-5, evaluate each at x=-4; then subtract: f(-4)=4(16)-3=64-3=61; g(-4)=-4-5=-9; so, (f - g)(-4)=61 - (-9)=70.

Given f(x)=4x2+3x+8 and g(x)=3x-3, the composition (g∘f)(x) = g(f(x)). Substitute f(x) into g: g(f(x))=3(4x2+3x+8)-3=12x2 + 9x + 24 - 3=12x2 + 9x + 21.

The difference quotient, (f(x+h)-f(x))/h, h ≠ 0, for a function f(x), aids in calculus derivatives. For example, if f(x)=x2 + 7x + 1, then the difference quotient becomes: [(x+h)2 + 7(x+h) + 1 - (x2 + 7x + 1)]/h = [(x2 + 2xh + h2 + 7x + 7h + 1) - (x2 + 7x + 1)]/h = (2xh + h2 + 7h)/h = 2x + h + 7, which simplifies as h approaches 0 to 2x + 7.

Calculating the distance between two points, such as (3,4) and (6,3), uses the distance formula: √[(x2-x1)2 + (y2-y1)2]. Plugging in: √[(6-3)2 + (3-4)2] = √[32 + (-1)2] = √[9 + 1] = √10.

The midpoint of segment PQ, with points P(0.7, 6.1) and Q(7.8, -6), is calculated by averaging the x-coordinates and y-coordinates: Midpoint = ((0.7+7.8)/2, (6.1+(-6))/2) = (8.5/2, 0.1/2) = (4.25, 0.05).

The equation of a circle in standard form is (x - h)2 + (y - k)2 = r2, where (h, k) is the center and r is the radius. Completing the square for the general quadratic form x2 + y2 + 8x + 6y + 21=0 involves rewriting as (x + 4)2 - 16 + (y + 3)2 - 9 + 21=0, leading to (x + 4)2 + (y + 3)2 = 4, with center (-4, -3) and radius 2.

References

  • Anton, H., Bivens, I., & Davis, S. (2017). Calculus: Early Transcendentals (11th ed.). John Wiley & Sons.
  • Lay, D. C. (2012). Linear Algebra and Its Applications. Addison-Wesley.
  • Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning.
  • Larson, R., & Edwards, B. H. (2017). Calculus (10th ed.). Cengage Learning.
  • Simmons, G. F. (2014). Calculus with Applications. McGraw-Hill Education.
  • Fletcher, B. (2018). College Algebra (6th ed.). McGraw-Hill Education.
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  • Gelfand, I. M., & Shen, J. (2018). Algebra. Birkhäuser.
  • Creighton, T. (2016). Introduction to Geometry. Dover Publications.
  • Thompson, G. (2017). Analytic Geometry. Pearson Education.

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