Given an equation, how do you figure out which tangent line has the greatest slope?

i'm given this equation (x^2+1)^ -1





(yes that is raised to negative one)





the question is where on the graph is the slope of the tangent line the greatest?





idk how to figure it out =x please help!Given an equation, how do you figure out which tangent line has the greatest slope?
Are you looking for the greatest slope, or the steepest slope? The slope can be very steep but negative.





y = (x虏+1)鈦宦?





A sketch of the graph shows that the slope is steepest at x=0, and flattens out as x moves away from 0:


http://www.flickr.com/photos/dwread/3690鈥?/a>





You can verify this algebraically:


slope = y'


= -(x虏+1)鈦宦?2x)


= -(2x)/(x虏+1)虏


The smaller the denominator, the steeper the slope. Thus x=0 has the greatest slope.Given an equation, how do you figure out which tangent line has the greatest slope?
The first derivative gives the slope. When the first derivative is zero, you can determine the maxima and minima of the function.





The second derivative gives the slope of the slope. Where the second derivative is zero, you can determine the maximum value of the slope.





f(x) = 1/(x虏 + 1)





f '(x) = - 2x/(x虏 + 1)虏





By the quotient rule,





f ''(x) = (6x虏 - 2)/(x虏 + 1)鲁





f ''(x) is maximum when (6x虏 - 2) = 0





6x虏 - 2 = 0





6x虏 = 2





x虏 = 1/3





x = +/- 1/sqrt(3)





The first derivative at one of these values is the maximum slope of f (x).








f '(x) = - 2x/(x虏 + 1)虏





f '(1/sqrt(3)) = - 2/[sqrt(3)][(sqrt(3))虏 + 1]虏





f ' = - 2/[sqrt(3)][3 + 1]虏





f ' = - 2/4sqrt(3)





f '(1/sqrt(3)) is negative. The tangent at - 1/sqrt(3) will be positive and the maximum value of the slope of the function.
It is at x = -1/sqrt(3).