[PRINT] MATH-1720 F23-04 , Question1: Score 3/3 Find horizontal asymptote of the curve y = 4 x 92 + 55 x 91 − 11 21 x 92 + 95 x 90 + 90 Note: Give the exact answer but not the decimal approximation(for example, write 4/5 istead of 0.8). Answer: The horizontal asymptote is y = Your response Correct response 4/21 Auto graded Grade: 1/1.0 A+ 100% Total grade: 1.0×1/1 = 100% Feedback: lim x → ∞ y = lim x → ∞ 4 x 92 + 55 x 91 − 11 21 x 92 + 95 x 90 + 90 = lim x → ∞ 4 x 92 + 55 x 91 − 11 x 92 21 x 92 + 95 x 90 + 90 x 92 = lim x → ∞ 4 + 55 x − 11 x 92 21 + 11 x 2 + 90 x 92 = 4 21 So the horizontal asymptote is y = 4 21 lim x → − ∞ y = lim x → − ∞ 4 x 92 + 55 x 91 − 11 21 x 92 + 95 x 90 + 90 = lim x → − ∞ 4 x 92 + 55 x 91 − 11 x 92 21 x 92 + 95 x 90 + 90 x 92 = lim x → − ∞ 4 + 55 x − 11 x 92 21 + 11 x 2 + 90 x 92 = 4 21 So there is only one horizontal asymptote y = 4 21  

Question2: Score 3/3 Find the horizontal asymptotes of the curve y = 87 e 16 x + 88 e − 77 x + 70 16 e 16 x − 17 e − 77 x + 65 Note: Give the exact answer but not the decimal approximation(for example, write 4/5 istead of 0.8). Answer: The equation of the horizontal asymptote which is above the x -axis is y = Your response Correct response 87/16 Auto graded Grade: 1/1.0 A+ 100% The equation of the horizontal asymptote which is below the x -axis is y = Your response Correct response -88/17 Auto graded Grade: 1/1.0 A+ 100% Total grade: 1.0×1/2 + 1.0×1/2 = 50% + 50% Feedback: lim x → ∞ y = lim x → ∞ 87 e 16 x + 88 e − 77 x + 70 16 e 16 x − 17 e − 77 x + 65 We will divide the denominator and denominator by e 16 x and obtain lim x → ∞ y = lim x → ∞ 87 + 88 e − 93 x + 70 e − 16 x 16 − 17 e − 93 x + 65 e − 16 x = 87 16 Since the answer of the limit is positive, the horizontal asymptote above the x − axis is y = 87 16 Now we will find lim x → − ∞ y = lim x → − ∞ 87 e 16 x + 88 e − 77 x + 70 16 e 16 x − 17 e − 77 x + 65 Dividing the denominator and denominator by e − 77 x we obtain lim x → − ∞ y = lim x → − ∞ 87 e 93 x + 88 + 70 e 77 x 16 e 93 x − 17 + 65 e 77 x = − 88 17 Since the answer of the limit is negative, the horizontal asymptote below the x − axis is y = − 88 17 Question3: Score 3/3   

Find the vertical asymptote of the curve y = 100 e 25 x + 31 e − 23 x 16 e 25 x − 9 e − 23 x Note: Give the exact answer but not the decimal approximation(for example, write 4/5 istead of 0.8). Write ln for natural logarithm not log_e. also remember to put brackets, for example, write ln(1/2) not ln 1/2. Answer: The vertical asymptote is x = Your response Correct response ln(e) (1/48)*ln(9/16) Auto graded Grade: 0/1.0 F 0% Total grade: 0.0×1/1 = 0% Feedback: We set the denomintor equal to zero and solve for x 16 e 25 x − 9 e − 23 x = 016 e 25 x = 9 e − 23 x e 48 x = 9/1648 x = ln(9/16) x = 1 48 ln 9 16 So x = 1 48 ln 9 16 is a possible vertical asymptote. Now we check the limit lim x → 1 48 ln 9 16 + y = lim x → 1 48 ln 9 16 + 100 e 25 x + 31 e − 23 x 16 e 25 x − 9 e − 23 x = ∞ So x=\frac{1}{48}\ln\left(\frac{9}{16}\right) is a vertical asymptote. Alternatively, we can check the left hand limit. \displaystyle \lim_{x\rightarrow \frac{1}{48}\ln\left(\frac{9}{16}\right)^-} y = \displaystyle \lim_{x\rightarrow \frac{1} {48}\ln\left(\frac{9}{16}\right)^-}\frac{100 e^{25 x}+31 e^{-23 x}}{16 e^{25 x}-9 e^{-23 x}} = - \infty So x=\frac{1}{48}\ln\left(\frac{9}{16}\right) is a vertical asymptote. Question5: Score 0/3 Find the values of 𝑐 and 𝑑 that makes the following function continuous on \mathbb{R} \displaystyle f(x) =\left\{\begin{array} {cl} c +d\left(\frac{x^2-64}{x-8}\right) & \textrm{if } x < 8 \\ 13 c (x- 8)^{38}\cos\left(\frac{3}{x-8}\right) +8 & \textrm{if } x > 8 \\ c x+ d & \textrm{if } x = 8 \end{array}\right. Note: Give the exact answer but not the decimal approximation(for example, write 4/5 istead of 0.8). Answer: The values of 𝑐 and 𝑑 are 𝑐 =   ( ) ( ) ( ) ( )

Your response Correct response 3 120/127 Auto graded Grade: 0/1.0 F 0% 𝑑 = Your response Correct response 2 56/127 Auto graded Grade: 0/1.0 F 0% Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0% Feedback: The function is a piecewise function and it is continuous for x< 8 and x> 8 . For f to be continuous on \mathbb{R} , it must also be continuous at x=8. So we must have \displaystyle \lim_{x\rightarrow 8^-} f(x)= \lim_{x\rightarrow 8^+} f(x) = f(8) \qquad \qquad (1) We calculate these values \begin{align*} \lim_{x\rightarrow 8^-}\ c +d\left(\frac{x^2-64}{x-8}\right)&=c +d \lim_{x\rightarrow 8^-} \left(\frac{(x-8)(x+8)} {x-8}\right)\\ &=c +d \lim_{x\rightarrow 8^-} (x+8)=c+16 d \end{align*} \displaystyle \lim_{x\rightarrow 8^+}\ 13 c (x-8)^{38}\cos\left(\frac{3}{x-8}\right) +8=13 c \lim_{x\rightarrow 8^+}\ (x- 8)^{38}\cos\left(\frac{3}{x-8}\right) +8 We know that -1\le \cos\left(\frac{3}{x-8}\right)\le1 -(x-8)^{38}\le (x-8)^{38} \cos\left(\frac{3}{x-8}\right)\le (x-8)^{38} Now \displaystyle\lim_{x\rightarrow 8^+} -(x-8)^{38}=0 and \displaystyle\lim_{x\rightarrow 8^+} (x-8)^{38}=0 So by the squeeze theorem \displaystyle \lim_{x\rightarrow 8^+}\ (x-8)^{38}\cos\left(\frac{3}{x-8}\right)=0 Therefore \displaystyle \lim_{x\rightarrow 8^+}\ 13 c (x-8)^{38}\cos\left(\frac{3}{x-8}\right) +8=13 c \lim_{x\rightarrow 8^+}\ (x-8)^{38}\cos\left(\frac{3}{x-8}\right) +8=8 And the value of the function at 8 is f(8)=8 c+d Substituting these values in equation (1) we obtain c+16 d=8=8 c+d So we have to solve the system of equations c+16 d=8 8 c+d=8 Solving the above system we get c=\frac{120}{127} d=\frac{56}{127}   

f'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{7(x^2-(x+h)^2)}{h x^2(x+h)^2} Auto graded Grade: 1/1.0 A+ 100% (iv) Which one is the correct statement (the next step to find 𝑓 ′ 𝑥 by definition)? Your response Correct response f'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{-7(2x+h)}{ x^2(x+h)^2} Auto graded Grade: 1/1.0 A+ 100% Total grade: 1.0×1/4 + 1.0×1/4 + 1.0×1/4 + 1.0×1/4 = 25% + 25% + 25% + 25% Feedback: \begin{align*} f'(x) &=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} \\ &=\lim_{h\rightarrow 0}\frac{\frac{7}{(x+h)^2}-\frac{7} {x^2}}{h} \\ &=\lim_{h\rightarrow 0}\frac{7 (x^2-(x+h)^2)}{h x^2(x+h)^2} \\ &= \lim_{h\rightarrow 0}\frac{-7 h(2x+h)}{ hx^2(x+h)^2}\\ &= \lim_{h\rightarrow 0}\frac{-7(2x+h)}{ x^2(x+h)^2}\\ &= \frac{-7(2x+0)}{ x^2(x+0)^2}\\ &= \frac{-14}{x^3} \end{align*} Please note that the last part is not asked in the question. Question8: Score 4/4 Let f(x)=\sqrt{x+28} . We will use the definition of derivative to find the derivative of 𝑓𝑥 . (i) What is the definition of derivative of 𝑓𝑥 ? Your response Correct response 𝑓 ′ 𝑥 = lim ℎ → 0 𝑓𝑥 + ℎ − 𝑓𝑥 ℎ Auto graded Grade: 1/1.0 A+ 100% (ii) Which one is the correct statement (the next step to find 𝑓 ′ 𝑥 by definition)? Your response Correct response \displaystyle f'(x)=\lim_{h\rightarrow 0} \frac{\sqrt{x+h+28}- \sqrt{x+28}}{h} Auto graded Grade: 1/1.0 A+ 100% (iii) Which one is the correct statement (the next step to find 𝑓 ′ 𝑥 by definition)? Your response Correct response f'(x)=\displaystyle \lim_{h\rightarrow 0} \frac{h} {h(\sqrt{x+h+28}+\sqrt{x+28})} Auto graded Grade: 2/2.0 A+ 100% (iv) Which one is the correct statement (the next step to find 𝑓 ′ 𝑥 by definition)? Your response Correct response      

f'(x)=\displaystyle \frac{1}{2\sqrt{x+28}} Auto graded Grade: 2/2.0 A+ 100% Total grade: 1.0×1/6 + 1.0×1/6 + 1.0×2/6 + 1.0×2/6 = 17% + 17% + 33% + 33% Feedback: f(x)=\sqrt{x+28} The derivative by definition is \begin{align*} f^\prime(x) &= \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h\rightarrow 0} \frac{\sqrt{x+h+28}- \sqrt{x+28}}{h} \\ &= \lim_{h\rightarrow 0} \frac{\sqrt{x+h+28}-\sqrt{x+28}}{h} \left(\frac{\sqrt{x+h+28}+\sqrt{x+28}} {\sqrt{x+h+28}+\sqrt{x+28}}\right) \\ &= \lim_{h\rightarrow 0} \frac{(x+h+28)-(x+28)}{h(\sqrt{x+h+28}+\sqrt{x+28})} \\ &= \lim_{h\rightarrow 0} \frac{h}{h(\sqrt{x+h+28}+\sqrt{x+28})} \\ &= \lim_{h\rightarrow 0} \frac{1} {\sqrt{x+h+28}+\sqrt{x+28}} = \frac{1}{2\sqrt{x+28}} \end{align*}  