This is a continuous from my last question
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.
On what intervals is the graph of g concave down? Justify your answer.
It looks like its between 1 and 3 but I do not know the exact points
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.
well, you can find the exact interval: it's where g" < 0
Since g" = (x^2-4x+16)/(x-2)^2
g" < 0 where x^2-4x+16 < 0
Since that is never negative, there is no interval where g(x) is concave down.
g(x) = (x^2+4x-13)/2 - 12log(x-2)
See that this is so at
http://www.wolframalpha.com/input/?i=%28x%5E2%2B4x-13%29%2F2+-+12log%28x-2%29
As for the tangent line at (3,4), it has slope g'(3) = -7, so it is
g-4 = -7(x-3)
Since g is concave up everywhere, the tangent must lie below the curve.
Plot that with g(x), and you have
http://www.wolframalpha.com/input/?i=plot+g%3D%28x%5E2%2B4x-13%29%2F2+-+12log%28x-2%29%2C+g%3D-7x%2B25
To determine the intervals where the graph of g is concave down, we need to examine the second derivative of g. The concavity of a function is determined by its second derivative: if the second derivative is negative, the function is concave down.
The given derivative of g is g′(x) = (x^2 – 16)/(x – 2).
To find the second derivative, we differentiate g′(x) using the quotient rule:
g''(x) = [(x – 2)((2x) – (x^2 – 16)) - (x^2 – 16)(1)] / (x – 2)^2
Simplifying this expression, we get:
g''(x) = (16 - 2x^2) / (x - 2)^2
To determine where g is concave down, we need to find where g''(x) < 0.
Setting g''(x) < 0:
(16 - 2x^2) / (x - 2)^2 < 0
Next, we determine the values of x that satisfy this inequality. To do this, we can create a sign chart by examining the sign of the numerator and denominator for different intervals. However, since the quadratic equation in the numerator is easily factorizable, we can solve it directly and find its roots.
Setting the numerator to zero:
16 - 2x^2 = 0
Factorizing:
2(x + 2)(x - 2) = 0
This equation yields two roots: x = -2 and x = 2. However, since x ≠ 2 according to the given conditions, the only relevant root is x = -2.
Now, we can analyze the intervals on the number line:
For x < -2, both the numerator and denominator are negative, so g''(x) > 0.
For -2 < x < 2 (excluding x = 2), the numerator is negative, and the denominator is positive, so g''(x) < 0. This is the interval where g is concave down.
For x > 2, both the numerator and denominator are positive, so g''(x) > 0.
Thus, the graph of g is concave down for -2 < x < 2.
To find the equation of the tangent line to the graph of g at x = 3, we need to find the value of g(3) and the slope of the tangent line, which is given by g'(3).
From the given information, we know that g(3) = 4.
To find g'(3), we substitute x = 3 into the derivative equation:
g'(3) = (3^2 – 16)/(3 - 2) = -7
Therefore, the slope of the tangent line is -7.
Using the point-slope form of a linear equation, we can write the equation of the tangent line:
y - y1 = m(x - x1), where (x1, y1) is the point on the curve (3, 4).
Substituting the values, we get:
y - 4 = -7(x - 3)
Expanding and rearranging, the equation becomes:
y = -7x + 25
So, the equation of the tangent line to the graph of g at x = 3 is y = -7x + 25.
To determine if the tangent line lies above or below the graph at this point, we compare the value of g(3) to the value of y on the tangent line when x = 3.
From before, we know that g(3) = 4.
Substituting x = 3 into the equation of the tangent line, y = -7x + 25:
y = -7(3) + 25
y = -21 + 25
y = 4
As we can see, the value of y on the tangent line when x = 3 matches the value of g(3) at the same x-coordinate. This indicates that the tangent line lies on the graph of g at that point, or in other words, the tangent line intersects the graph at x = 3. Therefore, the tangent line lies above the graph at this point.