b) Using the mean value theorem gave me f'(c)=1 and solving for c gave me 1 as well.
c) Tangent line y=x-13
I checked with a graphing calculator and my answers look right
Consider the graph of the function f(x)=x^2-x-12
a) Find the equation of the secant line joining the points (-2,-6) and (4,0).
I got the equation of the secant line to be y=x-4
b) Use the Mean Value Theorem to determine a point c in the interval (-2,4) such that the tangent line at c is parallel to the secant line.
I took the derivative of the original question to be f(x)=2x-1 but don't know what to do from there
c) Find the equation of the tangent line through c.
2 answers
your secant equation is correct
checking the other parts:
f '(x) = 2x - 1 ---> you had that
setting that equal to 1, since the slope of our secant is 1
2x-1 = 1
2x = 2
x = 1
so all you have to do is sub that back into the original
f(1) = 1 - 1 - 12 = -12
So the point where the slope of the secant is equal to the slope of the tangent is (1, -12)
the equation of the tangent at that point is
y+12 = 1(x-1)
y = x - 13 ----> you had that
good job.
You seem to be able to take care of the mechanics of the problem, but perhaps understanding what all that means seems a bit fuzzy.
I suggest you take a look at KhanAcademy video of this, especially the first two parts.
(These videos done by Sal Khan are extremely well done)
https://www.khanacademy.org/math/ap-calculus-ab/derivative-applications-ab/mean-value-theorem-ab/v/mean-value-theorem-1
The second part fits your problem extremely close
checking the other parts:
f '(x) = 2x - 1 ---> you had that
setting that equal to 1, since the slope of our secant is 1
2x-1 = 1
2x = 2
x = 1
so all you have to do is sub that back into the original
f(1) = 1 - 1 - 12 = -12
So the point where the slope of the secant is equal to the slope of the tangent is (1, -12)
the equation of the tangent at that point is
y+12 = 1(x-1)
y = x - 13 ----> you had that
good job.
You seem to be able to take care of the mechanics of the problem, but perhaps understanding what all that means seems a bit fuzzy.
I suggest you take a look at KhanAcademy video of this, especially the first two parts.
(These videos done by Sal Khan are extremely well done)
https://www.khanacademy.org/math/ap-calculus-ab/derivative-applications-ab/mean-value-theorem-ab/v/mean-value-theorem-1
The second part fits your problem extremely close
2 answers