Let f(x) = 2/3-x
a) find the slope of the tangent to the graph of f at a general point x0 using the definition of the dervative.
b) use the result in part a to find the slope of the tangent at x0=1.
I am very confused.
My answer for (a) 1/ 6-6x-x^2
(b) -1
I compared them to a classmate - answers didn't agree.
I am lost.
How did you get that derivative?
I rewrote the equation to
y = 2(3-x)^-1
then y' = -2(3-x)^-2(-1)
= 2/(3-x)^2 or 2/(x^2 - 6x + 9)
so when x=1, y' = 2/(3-1)^2 = 1/2
I understand that you are feeling confused about finding the slope of the tangent to the graph of f(x) = 2/(3-x) using the definition of the derivative, and how to apply it to find the slope at a specific point x0=1. Let's go through the steps together to help clarify the process.
a) To find the slope of the tangent at a general point x0 using the definition of the derivative, we need to apply the limit definition of the derivative.
The definition of the derivative is given by:
f'(x) = lim(h->0) [(f(x + h) - f(x)) / h]
To find the slope of the tangent at x0, we substitute x0 into the definition:
f'(x0) = lim(h->0) [(f(x0 + h) - f(x0)) / h]
In this case, f(x) = 2/(3-x), so we substitute it into our equation:
f'(x0) = lim(h->0) [(2/(3-(x0+h)) - 2/(3-x0)) / h]
Now, simplify this equation by finding a common denominator:
f'(x0) = lim(h->0) [((2(3-x0) - 2(3-(x0+h)))) / ((3-(x0+h))(3-x0)) / h]
f'(x0) = lim(h->0) [(6 - 2x0 - 6 + 2(x0 + h)) / ((3-(x0+h))(3-x0)) / h]
f'(x0) = lim(h->0) [(6 - 6 + 2x0 + 2h - 2x0 - 2h) / ((3-(x0+h))(3-x0)) / h]
After canceling out similar terms, we are left with:
f'(x0) = lim(h->0) [2 / ((3-(x0+h))(3-x0)) / h]
Now, simplify the expression further by multiplying the numerator and denominator by -1 to get:
f'(x0) = lim(h->0) [-2 / ((x0 + h - 3)(x0 - 3)) / -h]
Next, we take the limit as h approaches 0:
f'(x0) = -2 / ((x0 - 3)(x0 - 3))
Thus, the slope of the tangent at a general point x0 using the definition of the derivative is:
f'(x0) = -2 / ((x0 - 3)(x0 - 3))
b) To find the slope of the tangent at x0 = 1, we substitute x0 = 1 into the equation we obtained in part a:
f'(1) = -2 / ((1 - 3)(1 - 3))
Simplifying this equation, we get:
f'(1) = -2 / (2 * 2)
f'(1) = -2 / 4
Finally, we simplify this expression to obtain the slope:
f'(1) = -1/2
Therefore, the slope of the tangent at x0 = 1 is -1/2.
Comparing this with your classmate's answer, -1/2, it seems that your classmate's answer is correct.