Approximation Problems (Finding an equation of a Tangent Line)

Click For Summary
SUMMARY

The discussion focuses on finding the equation of the tangent line for the function f(x) = 3x² - 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 accuracy of this equation and guide the user to evaluate function values at f(x ± 0.01) and corresponding tangent line values. The user initially miscalculates these values but corrects them after feedback, demonstrating the importance of careful evaluation in calculus.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives
  • Familiarity with the point-slope formula for linear equations
  • Basic knowledge of evaluating functions
  • Ability to perform calculations with small increments (delta x)
NEXT STEPS
  • Learn about calculating derivatives for polynomial functions
  • Study the concept of limits and their application in calculus
  • Explore the use of Taylor series for approximating functions
  • Practice evaluating function values using different delta x values
USEFUL FOR

Students studying calculus, educators teaching derivatives and tangent lines, and anyone looking to improve their problem-solving skills in mathematical analysis.

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