how to calculate the tangent of a linear equation

Limits, differentiation, related rates, integration, trig integrals, etc.
jon80
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how to calculate the tangent of a linear equation

Postby jon80 » Sun Sep 30, 2012 4:55 pm

Is it correct to deduce the tangent of f(x) = 5x^3 + 5x^2 + x + 1 at x = 0.5 to be y = 9.75x - 1.5?

If yes, could you show the workings? I have this software named Graph v4.4 (http://www.padowan.dk), which deduced the tangent above, and, according to my knowledge the tangent would be calculated by drawing the dy/dx as follows:
15x^2 + 10x + 1 where, x = 0.5.

Am I mistaken?

nona.m.nona
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Re: how to calculate the tangent of a linear equation

Postby nona.m.nona » Mon Oct 01, 2012 4:19 am

jon80 wrote:Is it correct to deduce the tangent of f(x) = 5x^3 + 5x^2 + x + 1 at x = 0.5 to be y = 9.75x - 1.5?

If yes, could you show the workings? I have this software named Graph v4.4 (http://www.padowan.dk), which deduced the tangent above, and, according to my knowledge the tangent would be calculated by drawing the dy/dx as follows:
15x^2 + 10x + 1 where, x = 0.5.

Am I mistaken?

Lacking experience with your software and not knowing how you are using it, it is not possible to provide a review of its output. However, since tangent lines are typically linear equations, it seems unlikely that the equation of the tangent in a quadratic. Perhaps you are confusing "derivative of the equation" with "slope of the tangent at one particular point".

To determine the equation of the line which is tangent to a particular point, one finds the derivative, evaluates to find the value of the slope at that particular point, and then creates the linear equation with that slope at the given point, using methodology from one's earlier algebra coursework. Note that, since the derivative at that point is 15/4 + 10/2 + 1 = 39/4 = 9.75, the listed equation seems to be headed in the right direction.


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