1x2 at--1, make two tables of at least four values each For th showing slopes of

Question

1x2 at--1, make two tables of at least four values each For th showing slopes of secant lines that will help you to determine the slope of the tangent line at the given value of x. The first table should be values approaching x from the left and the second table should be values approachingx from the right. Label each table. (20) e function f(x) = a. Draw tables. left 2 ? ? lim 30769 .!?5 |-30769 .??9qS .kool V.LAaa S Using the results of part (a), guess the value of the slope of the tangent line to the curve at x-1 2. Slope of the tangent line: Using the slope from part (b), find an equation of the tangent line to the curve at Equation of the tangent line:

Explanation / Answer

1.

When we approach x = -1 from the left then some random value of x could be :

x = -1.01 , x = -1.001 , x = -1.0001 and x = -1.00001

Lets find the value of f(x) at these x - values

=> f(-1.01) = 1/[1 + (-1.01)^2] = 0.49502

=> f(-1.001) = 1/[1 + (-1.001)^2] = 0.4995002

=> f(-1.0001) = 1/[1 + (-1.0001)^2] = 0.49995002

=> => f(-1.00001) = 1/[1 + (-1.00001)^2] = 0.499995

When we approach x = -1 from the right then some random value of x could be :

x = - 0.9 , x = - 0.99 , x = - 0.999 and x = - 0.9999

Lets find the value of f(x) at these x - values

=> f(-0.9) = 1/[1 + (-0.9)^2] = 0.552486

=> f(- 0.99) = 1/[1 + (- 0.99)^2] = 0.50502

=> f(- 0.999) = 1/[1 + (- 0.999)^2] = 0.5005

=> f(- 0.9999) = 1/[1 + (- 0.9999)^2 = 0.50005

Looking at the two values of the function f(x) , the slope of the secant line is , msecant = 0.5

As the tangent line to the function f(x) = 1/(1+x^2) at x = - 1 would touch the function at x = -1.

And at x = -1 we know that slpe of the secant .

Hence at x = -1 the slope of the tangent line would be same as that of the secant line

=> mtangent = msecant = 0.5

Now when x = -1 , y = 1/(1 + (-1)^2) = 1/2 = 0.5

Hence using the point slope form : (y-y1) = m(x-x1) , we could find the equation of the tangent line

=> y - 0.5 = 0.5(x + 1)

or y = 0.5x + 1   -------> This is the required equation of the tangent line at x = - 1

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