MHB Approximation Problems (Finding an equation of a Tangent Line)

Click For Summary
The discussion focuses on finding the equation of the tangent line for the function f(x) = 3x^2 - 1 at the point (2, 11). The correct tangent line equation is derived using the point-slope formula, resulting in y = 12x - 13. Participants confirm the correctness of this equation and guide the user to evaluate function values at f(x ± 0.01). After correcting initial calculation errors, the user finds the new values for the tangent line and function, which are now much closer together. The conversation emphasizes learning from mistakes and improving problem-solving skills.
mathkid3
Messages
23
Reaction score
0
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 !
 
Physics news on Phys.org
Your tanget line is correct.

The function values will be of the form f(x-0.01), can you continue?
 
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 $\displaystyle y(x\pm0.01)$.
 
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 = 23.12Is this right fellas?(Thinking)
 
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.
 
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 ? (Envy)
 
Yes, those values are correct!

Hey, it was a simple mistake, the kind most of us make from time to time. So don't be discouraged. You will get better at recognizing when you have made a mistake like this. Your clue this time was the fact that the values were so far apart.(Wave)
 

Similar threads

Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 53 ·
2
Replies
53
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K