MATHEMATICS 203 MIDTERM EXAM 2 QUESTIONS WITH WORKED OUT SOLUTIONS 100% CORRECT || VERIFIED MATH PAPER

MATHEMATICS 203 MIDTERM EXAM 2 QUESTIONS WITH WORKED OUT SOLUTIONS 100% CORRECT || VERIFIED MATH PAPER (11 marks):(a) An open rectangular box (no top) with volume 3 cubic meters has a square base. If the length of each side of the base is (x) and the height is (h), express the surface area (S) of the box as a function of (x) only (not (x) and (h)).(b) Let (f(x)=sqrt{2x-8}) and (g(x)=x^{2}-5) Find the composite functions (fcirc g) and (gcirc f), and determine their domains.(c) Find the inverse function (f^{-1}(x)) of (f(x)=log _{5}(2x-1)) and determine the domain and the range of (f(x)).(8 marks) Find the limit or explain why the limit does not exist:(a) (lim _{trightarrow 0}left(frac{1}{t}-frac{1}{tsqrt{1+t}}right))(b) (lim _{xrightarrow 3}frac{x|x-3|}{x^{2}-9})(4 marks) Find all horizontal and vertical asymptotes of the graph(y=frac{x^{2}-x}{x^{2}-6x+5}) Thank youYour feedback helps Google improve. See our Privacy Policy.Share more feedbackReport a problemClose 5. (16 marks) Find the derivatives of the following functions. (You don't need to simplify the final answer, but you must show how you calculate it): (a) f(x) = x * sqrt(x) * (x + 1/x) ^ 2 (b) f(x) = (1 + x ^ 3) * e ^ (3x) (c) f(x) = (cos^2 x)/(1 + tan x) (d) f(x) = sqrt(x ^ 2 + cos(e ^ (x ^ 3 * sin x))) e ^ (3x) + e ^ (3x) * x ^ 3 6. (6 marks) Given the function f(x) = (2x)/(2 + x) (a) Calculate f' * (x) using its definition as a limit of difference quotient. (b) Write equation of the tangent line to the curve y = f(x) at the point (2, f(2) ) Bonus Question (3 marks). Give an example of a function f(x) for which f' * (0) exists but f^ prime prime (0) does not, or explain why this is impossible. we multiplied the nominator as well as the denominator by the conjugate expression sqrt(1 + t) + 1 Then, lim t -> 0 (1/t - 1/(t * sqrt(1 + t))) = lim t -> 0 t/(t * sqrt(1 + t) * (sqrt(1 + t) + 1)) = lim t -> 0 1/(sqrt(1 + t) * (sqrt(1 + t) + 1)) = 1/2 (b) lim x -> 3 (x|x - 3|)/(x ^ 2 - 9) = lim x -> 3 (x(x - 3))/(x ^ 2 - 9) = lim x -> 3 x/(x + 3) = 1/2 and from the left hand limit and then, lim x -> 3 (x|x - 3|)/(x ^ 2 - 9) The limit from the right is different does not exist. lim x -> 3 (x|x - 3|)/(x ^ 2 - 9) = lim x -> 3 (- x * (x - 3))/(x ^ 2 - 9) = lim x -> 3 (- x)/(x + 3) = - 1/2 3. First of all, the domain of y = 1 is R{1,5}. Then the (Indeterminate form 5/0) and then, The line 5 is a lim x -> 5 (x ^ 2 - x)/(x ^ 2 - 6x + 5) = lim x -> 5 x/(x - 5) = ∞ lim x -> ∞ * (or + - ∞) (x ^ 2 - x)/(x ^ 2 - 6x + 5) = lim x -> ∞ (x(x - 1))/((x - 1)(x - 5)) = lim x -> ∞ x/(x - 5) = lim x -> ∞ x/(x - 5) = lim x -> ∞ 1/(1 - 5/x) = 1 y = (x ^ 2 - x)/(x ^ 2 - 6x + 5) 3