Let g be a function that is defined for all x, x ≠ 2, such that g(3)=4 and the derivative of g is g′(x)=
x^2–16/x−2, with x≠2.
Write an equation for the tangent line to the graph of g at the point where x = 3.
Does this tangent line lie above or below the graph at this point? Justify your answer.
I will assume you meant
if g ' (x) = x^2 - 16/(x - 2)
or else why would you have x ≠ 2
g(x) = (1/3)x^3 - 16ln(x-2) + c
but (3,4) lies on it, so
4 = (1/3)(27) - 16ln1 + c
4 = 9 - 0 + c
c = -5
so g(x) = (1/3)x^3 - 16ln(x-2) - 5
when x = 3
g ' (x) = 9 - 16/1 = -7
tangent equation:
y-4 = -7(x-3)
or y = -7x + 25
see graph:
http://www.wolframalpha.com/input/?i=plot+y+%3D+%281%2F3%29x%5E3+-+16ln%28x-2%29+-+5+%2C+y+%3D+-7x%2B25
On the other hand, if you meant
g'(x) = (x^2–16)/(x−2)
= x+2 - 12/(x-2)
then
g(x) = x^2/2 + 2x - 12log(x-2) + c
g(3) = 9/2 + 6 - 12log(1) + c = 4, so c=-13/2
g'(3) = -7
and the tangent line is
y-4 = -7(x-3)
and the graph is at
http://www.wolframalpha.com/input/?i=plot+y%3Dx^2%2F2+%2B+2x+-+12log%28x-2%29+-+13%2F2%2C+y+%3D+-7%28x-3%29%2B4
To find the equation of the tangent line to the graph of g at x = 3, we need two pieces of information: the slope of the tangent line and a point on the line.
The slope of the tangent line can be determined using the derivative of g. In this case, the derivative g′(x) is given as
g′(x) = x^2 - 16 / (x - 2)
We can substitute x = 3 into the derivative to find the slope of the tangent line at x = 3:
g′(3) = (3^2 - 16) / (3 - 2) = (9 - 16) / 1 = -7
So, the slope of the tangent line at x = 3 is -7.
To find a point on the tangent line, we can use the function g(3) = 4, which represents the y-coordinate of the point where the tangent line intersects the graph of g at x = 3. Therefore, the point (3, 4) lies on the tangent line.
Now that we have the slope of the tangent line (-7) and a point on the line (3, 4), we can use the point-slope form of the equation of a line:
y - y1 = m(x - x1)
Here, (x1, y1) is the known point on the line (3, 4), and m is the slope of the line (-7).
Substituting the known values into the equation, we have:
y - 4 = -7(x - 3)
Expanding and simplifying:
y - 4 = -7x + 21
y = -7x + 25
So, the equation of the tangent line to the graph of g at x = 3 is y = -7x + 25.
Now, let's determine whether the tangent line lies above or below the graph at this point. The graph of g is not given explicitly, but we can use the information provided to analyze it.
Since the derivative g′(x) is given as (x^2 - 16) / (x - 2), we can find the critical points by setting the derivative equal to zero:
(x^2 - 16) / (x - 2) = 0
The numerator can only equal zero when x^2 - 16 = 0, which occurs when x = -4 or x = 4. But the denominator cannot be zero, so x = 2 cannot be a critical point.
This means that the graph of g has no critical points between -4 and 4. Since the tangent line intersects the graph at (3, 4), and the slope is negative (-7), we can conclude that the tangent line lies above the graph at this point.
Alternatively, we can also observe that the slope of the tangent line is negative (-7). In general, the slope of the tangent line represents the rate at which the function is decreasing (or increasing) at that point. Since the slope is negative, it indicates that the function is decreasing. Therefore, the tangent line lies above the graph at x = 3.