How do you find the equations of the tangent to the curve #y = (1 - x)(1 + x^2)-1# that pass through the point (1, 2)?

Answer 1

Find the equation for the line tangent to the curve at #x=a#. Then substitute #x=1# and #y=2# to find #a#. The rewrite the tangent line using that value of #a#. I got #y = -7.95x+9.95# (approx).
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Simplify the function #y = f(x) = -x^3+x^2-x#
We shall find the equation of the line tangent to the curve at #(a,f(a))#.
#f(a) = -a^3+a^2-a#
#f'(x) = -3x^2+2x-1#, so the slope of the tangent line is
#m = f'(a) = -3a^2+2a-1#

The tangent line has equation:

#y-(-a^3+a^2-a) = (-3a^2+2a-1)(x-a)#

The slope-intercept form is

#y = (-3a^2+2a-1)x+(2a^3-a^2)#
we shall now find #a# to make the tangent line contain the point #(1,2)#
Solve for #a#
#2 = (-3a^2+2a-1)(1)+(2a^3-a^2)#
#2a^3-4a^2+2a-3 = 0#

Use whatever tools you have to solve this cubic. I got

#a ~~ 1.89#

So the equation of the tangent line is (approximately)

#y = -7.95x+9.95#
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Answer 2

To find the equations of the tangent to the curve y = (1 - x)(1 + x^2)^-1 that pass through the point (1, 2), we need to follow these steps:

  1. Differentiate the given curve equation with respect to x to find the derivative dy/dx.
  2. Substitute the x-coordinate of the given point (1, 2) into the derivative to find the slope of the tangent line at that point.
  3. Use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope found in step 2.
  4. Simplify the equation obtained in step 3 to find the equation of the tangent line.

Let's go through these steps:

  1. Differentiating y = (1 - x)(1 + x^2)^-1: dy/dx = -[(1 + x^2)^-1] + (1 - x)(-1)(2x)(1 + x^2)^-2 Simplifying: dy/dx = -[(1 + x^2) + 2x^2(1 - x)] / (1 + x^2)^2

  2. Substituting x = 1 into the derivative: dy/dx = -[(1 + 1^2) + 2(1)^2(1 - 1)] / (1 + 1^2)^2 Simplifying: dy/dx = -2/4 = -1/2

  3. Using the point-slope form with (x1, y1) = (1, 2) and m = -1/2: y - 2 = (-1/2)(x - 1)

  4. Simplifying the equation: y - 2 = (-1/2)x + 1/2 y = (-1/2)x + 5/2

Therefore, the equation of the tangent line to the curve y = (1 - x)(1 + x^2)^-1 that passes through the point (1, 2) is y = (-1/2)x + 5/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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