Analyzing functions #4 - Questions and Answers

Question 1 Describe the First Derivative Test in your own words. Answer The first derivative test is the first step in the process of studying functions to determine their extreme point. The point is a crucial component of the function. x = c at which either f'(c) = 0 or f'(c) does not exist. A function is said to have a local (relative) extreme at a crucial point if the derivative of the function changes sign near that location. At the critical point, the function has a local (relative) maximum if the derivative switches from being positive (growing function) to negative (decreasing function). The function has a local (relative) minimum at the critical point if the derivative switches from a negative (decreasing function) to a positive (growing function). The First Derivative Test for Local Extreme Values is the name of the procedure when it is used to find the local maximum or lowest function values. Step 2 For example, Let, f(x) = x⁴ - 8x² Differentiate with respect to x, f'(x) = 4x³ - 16x Find critical points by solving f'(x) = 0 for x, f'(x) = 0 4x³ - 16x = 0 4x(x² - 4) = 0 4x(x - 2)(x + 2) = 0 x = 0, x = 2, x = -2 Step 3 f(0) = (0)⁴ - 8(0)² = 0 f(2) = (2)⁴ - 8(2)² = -16 f(-2) = (-2)⁴ - 8(-2)² = -16 Consequently, the function's key components are (0, 0), (2, -16) and (-2, -16). The domain of the function (-∞, ∞) is split into four smaller periods, (-∞, -2), (-2, 0), (0, 2) and (2, ∞) Step 4 Interval Test point f'(x) Increasing/Decreasing (-∞, -2) x = -3 f'(-3) = 4(-3)³ - 16(-3) = -60 < 0 Decreasing (-2, 0) x = -1 f'(-1) = 4(-1)³ - 16(-1) = 12 > 0 Increasing (0, 2) x = 1 f'(1) = 4(1)³ - 16(1) = -12 < 0 Decreasing (2, ∞) x = 3 f'(3) = 4(3)³ - 16(3) = 60 > 0 Increasing Because f'(x) changes from negative to positive around -2 and 2, f(x) has a local minimum at (-2, -16) and (2, -16). Also, f'(x) has a local maximum at 0 because its value varies from positive to negative at (0, 0). Question 2 By examining the appropriate limits, ascertain the final behavior of the following transcendental functions. The associated graph is then simply sketched, with any asymptotes indicated. f(x) = 2^x Answer Step 1 To determine the end behavior of the following transcendental functions by analyzing appropriate limits. Given function is f(x) = 2^x To find end behavior of function f(x) = 2^x we have to find asymptotes for f(x) = 2^x To find horizontal asymptotes: The line y = L is a horizontal asymptote of the function y = f(x) if either lim(x→∞) f(x) = L or lim(x→-∞) f(x) = L and L is finite. Now here, lim(x→∞) 2^x = 2^∞ = ∞ and lim(x→-∞) 2^x = 2^-∞ = 0 Therefore the line y=0 is the horizontal asymptote for f(x) = 2^x Step 2 To find vertical asymptotes: The line x = L is a vertical asymptote of the function f(x) = 2^x if the limit of the function at this point is infinite. Here the function f(x) = 2^x is defined for every value of x. Therefore there are no vertical asymptotes for f(x) = 2^x To find slant asymptotes: The line y = mx + b is slant asymptote of a function y = f(x) if either m = lim(x→∞) (f(x)/x) = L or m = lim(x→-∞) (f(x)/x) = L and L is finite and zero m = lim(x→∞) (2^x/x) = ∞ and m = lim(x→-∞) (2^x/x) = lim(x→-∞) (2^x/x) = lim(x→-∞) (1/(x2^-x)) = 0 Here the value of L is not finite and non zero. Therefore the function f(x) = 2^x has no slant asymptotes. Step 3 The sketch of the graph of function f(x) = 2^x is, Question 3 Calculus studies change by analyzing slopes of secant lines over successively shorter intervals.Determine whether the statement makes sense or does not make sense, and explain your reasoning. Answer Given statement is 'Calculus studies change by analysing slopes of secant lines over succesively shorter intervals' This sentence makes send because this is one of the topic of Calculus subject. This is one of the calculus subject's topics, so this sentence makes sense. To approximate the slope of the tangent line to the function graph at that particular point, find the secant line slopes over progressively shorter intervals. The slope of secant line on interval [a, a + h] is, (f(a+h)-f(a))/h The slope tangent line can be approximates by the slope of secant line by making h smaller and smaller. That means to find the slopes of secant line successively over the shorter intervals for smaller values of h. Therefore, the slope of tangent line m at point (a, f(a)) is m = lim(h→0) (f(a + h)-f(a))/h Question 4 List the various rules that allow us to differentiate constant functions, power functions, polynomials, exponential functions, rational functions, and certain combinations of them, simply and directly, without having to take limits each time? Answer Step 1 To explain: The differentiation rule: Explanation: The constant rule is used to differentiate constant function: The differentiation of constant function is 0. d/dx (k) = 0 The differentiation of power function rule. d/dx (kxⁿ) = k * d/dx (xⁿ) Step 2 Power rule of differentiation: Power rule of differentiation is used to differentiate polynomial function: d/dx (xⁿ) = n * xⁿ⁻¹ Exponential rule of differentiation: d/dx (eˣ) = eˣ Quotient rule of differentiation: Quotient rule of differentiation used to differentiate rational function: d/dx (u/v) = (vu' - uv')/v² Chain rule of differentiation: Chain rule of differentiation used for composite function. (f(g(x)))' = f'(g(x))g'(x) Thus, the every function has a rule for the differentiation. Question 5 What are the derivatives of the six basic hyperbolic functions? What are the corresponding integral formulas? What similarities do you see here to the six basic trigonometric functions? Answer Step 1 Derivatives of the six basic hyperbolic functions: d/dx (sinh x) = cosh x d/dx (cosh x) = sinh x d/dx (tanh x) = sech²x d/dx (sech x) = -sech x tanh x d/dx (csch x) = -csch x coth x d/dx (coth x) = -csch²x Step 2 Integrals of the six basic hyperbolic functions: ∫ sinh x dx = cosh x + C ∫ cosh x dx = sinh x + C ∫ tanh x dx = ln |cosh x| + C ∫ csch x dx = ln | tanh (x/2) | + C ∫ sech x dx = tan⁻¹ (sinh x) + C ∫ coth x dx = ln | sinh x | + C Similarities are the derivative and integral of sinh x & cosh x are same.