Approximation Problems (Finding an equation of a Tangent Line)

mathkid3

New member
I am asking for simple guidance on this problem.

f(x) = 3x^2-1, (2,11)

I do believe I need to obtain an equation for tan line so first step I think is to use point slope or slope intercept (a friendly reminder to the name of formula would be very nice )

y - ysub1 = m(x-xsub1)

= y - f(2) = f '(2)(x-2)

= y - 11 = 12(x-2) =

y = 12x -13

am I correct thus far obtaining the equation of the Tan Line for this specific problem?

also, what is my next step ? I am told after I get the equation for the tan line to then find the function values and the tan line values at f(x + delta x) and y(x+delta x) for delta x = -0.01 and 0.01

Thanks very much !

pickslides

Member
Your tanget line is correct.

The function values will be of the form f(x-0.01), can you continue?

MarkFL

Staff member
Yes, that is the correct tangent line. The formula you used is aptly named the point-slope formula, because it contains as parameters, the slope $\displaystyle m$ and the point $\displaystyle (x_1,y_1)$.

For the next step, evaluate:

$\displaystyle f(x\pm0.01)$ and $\displaystyle y(x\pm0.01)$.

mathkid3

New member
so like this...

f(1.99) = 3(1.99)-1 = 4.97
f(2.01) = 3(2.01)-1 = 5.03

and

y(1.99) = 12(1.99)-1 = 22.88
y(2.01) = 12(2.01) - 1 = 23.12

Is this right fellas?

MarkFL

Staff member
No, the reason the values are so far off, is that x is squared in the function definition and you need to subtract 13, not 1 in your tangent line. Try it again, and your values will be much closer.

mathkid3

New member
wow...I made the changes and they were easy changes I missed

the new values Mark is 10.8803,11.1203

y function values are as follows

10.88 and 11.12

Thanks Mark! What would I ever do without you? Think on my own ?