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.
Find all values of x where the graph of g has a critical value.
For each critical value, state whether the graph of g has a local maximum, local minimum or neither.
since g'=0 when x = ±4, those are the only two possibilities.
g' does not exist when x=2, but that does not matter here.
Since g" > 0 at both points, both are minima.
See the graph at
http://www.wolframalpha.com/input/?i=integral+%28x%5E2%E2%80%9316%29%2F%28x%E2%88%922%29+dx
To find the critical values of the function g(x), we need to determine where the derivative g'(x) is equal to zero or undefined.
First, let's find where g'(x) is equal to zero by setting the numerator equal to zero:
x^2 - 16 = 0
This is a basic quadratic equation, so we can factor it:
(x - 4)(x + 4) = 0
Setting each factor equal to zero, we get:
x - 4 = 0 or x + 4 = 0
Solving for x, we find two critical values:
x = 4 or x = -4
Now, let's check if these values are valid by determining if g'(x) is undefined at x = 2. Since the derivative is defined for all x except x = 2, we don't need to worry about that here.
Next, we need to determine the nature of the critical values. To do this, we analyze the intervals between the critical values. We can use a number line to help visualize this.
We have three intervals:
Interval 1: x < -4
Interval 2: -4 < x < 4
Interval 3: x > 4
Now, we can choose a test point within each interval and evaluate the sign of g'(x). Let's take x = -5 for Interval 1, x = 0 for Interval 2, and x = 5 for Interval 3.
For x = -5 (Interval 1):
Substitute x = -5 into g'(x):
g'(-5) = ((-5)^2 - 16)/(-5 - 2) = (25 - 16)/(-7) = 9/(-7) = -9/7
Since g'(-5) is negative, the graph of g has a local maximum at x = -5.
For x = 0 (Interval 2):
Substitute x = 0 into g'(x):
g'(0) = (0^2 - 16)/(0 - 2) = (-16)/(-2) = 8
Since g'(0) is positive, the graph of g has a local minimum at x = 0.
For x = 5 (Interval 3):
Substitute x = 5 into g'(x):
g'(5) = (5^2 - 16)/(5 - 2) = (25 - 16)/3 = 9/3 = 3
Since g'(5) is positive, the graph of g has a local minimum at x = 5.
To summarize:
The critical values of g(x) are x = 4 and x = -4.
The graph of g has a local maximum at x = -5.
The graph of g has a local minimum at x = 0 and x = 5.
Hope this helps! Let me know if you have any further questions.