Find all values for c in the intercal (1,4) so that the slope of the tangent line at (c,f(c)) equals the slope of the secant line through (1,1/3) and (4,2/3) where f(x) = x/(x+2) 1 </ x </ 4 (<-- that is suppose to be the less than or equal to sign)
I'm not sure how to do this. I think this involves the Mean Value Theorem but I'm not sure where I should begin.
This is indeed the MVT.
The secant line through (1,1/3) and (4,2/3) has slope 1/9. The MVT says that in the interval (1,4), there is some c where f'(c) = 1/9.
f'(x) = 2/(x+2)^2
So, we want c such that
2/(c+2)^2 = 1/9
(c+2)^2 = 18
c+2 = ±3√2
c = -2±3√2
We want c to be in (1,4), so that means we want c = -2+3√2 = 2.24
So, as the MVT says, there is such a c.
f(-2+3√2) = (-2+3√2)/(3√2) = 1 - √2/3
Thus, the tangent line is
y-(1-√2/3) = 1/9 (x-(-2+3√2))
The secant line is
y = 1/3 + 1/9 (x-1)
The graph of f(x) and the tangent and secant lines there are shown at
http://www.wolframalpha.com/input/?i=plot+y%3Dx%2F%28x%2B2%29%2C+y%3D+%281-%E2%88%9A2%2F3%29+%2B+1%2F9+%28x-%28-2%2B3%E2%88%9A2%29%29%2C+y+%3D+1%2F3+%2B+1%2F9+%28x-1%29+for+x%3D0..5
Oh wow, thank you so much!
To find the values of c in the interval (1,4) that satisfy the given condition, we can start by finding the slope of the tangent line at any point (c, f(c)) on the graph of f(x) = x/(x+2).
The slope of the tangent line can be determined by taking the derivative of the function f(x). Let's find the derivative of f(x) by applying the quotient rule:
f(x) = x/(x+2)
f'(x) = [(x+2)(1) - x(1)] / (x+2)^2
= (x + 2 - x) / (x+2)^2
= 2 / (x+2)^2
Now, we have the equation for the slope of the tangent line at any point (c, f(c)), which is f'(c) = 2 / (c+2)^2.
Next, we need to find the slope of the secant line through the points (1, 1/3) and (4, 2/3). The slope of a secant line is determined by the difference in the y-coordinates divided by the difference in the x-coordinates:
slope of secant line = (2/3 - 1/3) / (4 - 1)
= (1/3) / 3
= 1/9
Now, we can set up an equation to find the values of c that satisfy the condition:
f'(c) = slope of secant line
2 / (c+2)^2 = 1/9
To solve for c, we can cross-multiply:
2 * 9 = (c+2)^2
18 = (c+2)^2
Taking the square root of both sides:
√18 = c+2
c + 2 = ±√18
c = -2 ±√18
However, we are only interested in values of c within the interval (1,4). Therefore, we need to check which of these solutions lie within the specified interval:
-2 ± √18 ≈ -5.243 and -0.757
Since -5.243 is outside the interval (1,4), we can conclude that the only value of c that satisfies the given condition is approximately -0.757.