61 - 64. equations of tangent lines\na. find an equation of the line tangent to the given curve at a.\nb…

61 - 64. equations of tangent lines\na. find an equation of the line tangent to the given curve at a.\nb. use a graphing utility to graph the curve and the tangent line on the same set of axes.\n61. $y=\frac{x + 5}{x - 1};a = 3$\n62. $y=\frac{2x^{2}}{3x - 1};a = 1$\n63. $y = 1+2x+xe^{x};a = 0$\n64. $y=\frac{e^{x}}{x};a = 1$

61 - 64. equations of tangent lines\na. find an equation of the line tangent to the given curve at a.\nb. use a graphing utility to graph the curve and the tangent line on the same set of axes.\n61. $y=\frac{x + 5}{x - 1};a = 3$\n62. $y=\frac{2x^{2}}{3x - 1};a = 1$\n63. $y = 1+2x+xe^{x};a = 0$\n64. $y=\frac{e^{x}}{x};a = 1$

Answer

  1. For problem 61: (y=\frac{x + 5}{x - 1}), (a = 3)
    • Step - 1: Find the derivative of (y) using the quotient - rule
      • The quotient - rule states that if (y=\frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here, (u=x + 5), so (u^\prime=1), and (v=x - 1), so (v^\prime=1).
      • (y^\prime=\frac{1\cdot(x - 1)-(x + 5)\cdot1}{(x - 1)^{2}}=\frac{x - 1-x - 5}{(x - 1)^{2}}=\frac{-6}{(x - 1)^{2}}).
    • Step - 2: Find the slope of the tangent line at (x = a=3)
      • Substitute (x = 3) into (y^\prime): (y^\prime(3)=\frac{-6}{(3 - 1)^{2}}=\frac{-6}{4}=-\frac{3}{2}).
    • Step - 3: Find the (y) - coordinate when (x = 3)
      • Substitute (x = 3) into (y=\frac{x + 5}{x - 1}): (y(3)=\frac{3 + 5}{3 - 1}=\frac{8}{2}=4).
    • Step - 4: Use the point - slope form (y - y_1=m(x - x_1)) to find the equation of the tangent line
      • Here, (x_1 = 3), (y_1 = 4), and (m=-\frac{3}{2}).
      • (y - 4=-\frac{3}{2}(x - 3)).
      • Expand: (y-4=-\frac{3}{2}x+\frac{9}{2}).
      • Rearrange to the slope - intercept form (y=-\frac{3}{2}x+\frac{9}{2}+4=-\frac{3}{2}x+\frac{9 + 8}{2}=-\frac{3}{2}x+\frac{17}{2}).
  2. For problem 62: (y=\frac{2x^{2}}{3x - 1}), (a = 1)
    • Step - 1: Find the derivative of (y) using the quotient - rule
      • Let (u = 2x^{2}), then (u^\prime=4x), and (v=3x - 1), then (v^\prime=3).
      • (y^\prime=\frac{4x(3x - 1)-2x^{2}\cdot3}{(3x - 1)^{2}}=\frac{12x^{2}-4x - 6x^{2}}{(3x - 1)^{2}}=\frac{6x^{2}-4x}{(3x - 1)^{2}}).
    • Step - 2: Find the slope of the tangent line at (x = a = 1)
      • Substitute (x = 1) into (y^\prime): (y^\prime(1)=\frac{6\times1^{2}-4\times1}{(3\times1 - 1)^{2}}=\frac{6 - 4}{4}=\frac{1}{2}).
    • Step - 3: Find the (y) - coordinate when (x = 1)
      • Substitute (x = 1) into (y=\frac{2x^{2}}{3x - 1}): (y(1)=\frac{2\times1^{2}}{3\times1 - 1}=\frac{2}{2}=1).
    • Step - 4: Use the point - slope form (y - y_1=m(x - x_1)) to find the equation of the tangent line
      • Here, (x_1 = 1), (y_1 = 1), and (m=\frac{1}{2}).
      • (y - 1=\frac{1}{2}(x - 1)).
      • Expand: (y-1=\frac{1}{2}x-\frac{1}{2}).
      • Rearrange to the slope - intercept form (y=\frac{1}{2}x-\frac{1}{2}+1=\frac{1}{2}x+\frac{1}{2}).
  3. For problem 63: (y=1 + 2x+xe^{x}), (a = 0)
    • Step - 1: Find the derivative of (y) using the sum - rule and product - rule
      • The sum - rule: ((u + v+w)^\prime=u^\prime+v^\prime+w^\prime). Let (u = 1), (v = 2x), (w=xe^{x}).
      • (u^\prime = 0), (v^\prime=2). For (w = xe^{x}), use the product - rule ((uv)^\prime=u^\prime v+uv^\prime) where (u = x), (u^\prime = 1), (v = e^{x}), (v^\prime=e^{x}), so (w^\prime=e^{x}+xe^{x}).
      • (y^\prime=0 + 2+(e^{x}+xe^{x})=2+e^{x}(1 + x)).
    • Step - 2: Find the slope of the tangent line at (x = a = 0)
      • Substitute (x = 0) into (y^\prime): (y^\prime(0)=2+e^{0}(1 + 0)=2 + 1=3).
    • Step - 3: Find the (y) - coordinate when (x = 0)
      • Substitute (x = 0) into (y=1 + 2x+xe^{x}): (y(0)=1+2\times0+0\times e^{0}=1).
    • Step - 4: Use the point - slope form (y - y_1=m(x - x_1)) to find the equation of the tangent line
      • Here, (x_1 = 0), (y_1 = 1), and (m = 3).
      • (y - 1=3(x - 0)).
      • So (y=3x + 1).
  4. For problem 64: (y=\frac{e^{x}}{x}), (a = 1)
    • Step - 1: Find the derivative of (y) using the quotient - rule
      • Let (u = e^{x}), then (u^\prime=e^{x}), and (v = x), then (v^\prime=1).
      • (y^\prime=\frac{e^{x}\cdot x-e^{x}\cdot1}{x^{2}}=\frac{e^{x}(x - 1)}{x^{2}}).
    • Step - 2: Find the slope of the tangent line at (x = a = 1)
      • Substitute (x = 1) into (y^\prime): (y^\prime(1)=\frac{e^{1}(1 - 1)}{1^{2}}=0).
    • Step - 3: Find the (y) - coordinate when (x = 1)
      • Substitute (x = 1) into (y=\frac{e^{x}}{x}): (y(1)=\frac{e^{1}}{1}=e).
    • Step - 4: Use the point - slope form (y - y_1=m(x - x_1)) to find the equation of the tangent line
      • Here, (x_1 = 1), (y_1 = e), and (m = 0).
      • (y - e=0(x - 1)).
      • So (y = e).

Answer:

  • For problem 61: The equation of the tangent line is (y=-\frac{3}{2}x+\frac{17}{2}).
  • For problem 62: The equation of the tangent line is (y=\frac{1}{2}x+\frac{1}{2}).
  • For problem 63: The equation of the tangent line is (y=3x + 1).
  • For problem 64: The equation of the tangent line is (y = e).