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. You must justify your answers with a complete sentence.
On what intervals is the graph of g concave down? Justify your answer.
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.
g has a critical value where g'=0: x = ±4
max/concave down if g" < 0
g" = (x^2-4x+16)/(x-2)^2
For the tangent line, you have a point (3,4) and a slope (-7), so the line is
y-4 = -7(x-3)
To find the critical values of the function g(x), we need to find the values of x where the derivative g'(x) is equal to zero or is undefined (discontinuity).
1. Set the derivative equal to zero and solve for x:
(x^2 – 16)/(x – 2) = 0
(x + 4)(x – 4)/(x – 2) = 0
Setting each factor equal to zero separately:
(x + 4) = 0 --> x = -4
(x - 4) = 0 --> x = 4
These are the two critical values of the function g(x).
2. To determine whether the graph of g has a local maximum, local minimum, or neither at each critical value, we can use the second derivative test. The second derivative is obtained by differentiating the expression g'(x):
g''(x) = (2x(x - 4) - (x^2 - 16))/((x - 2)^2)
Simplifying the expression, we get:
g''(x) = 8/(x - 2)^2
Substituting each critical value into g''(x):
g''(-4) = 8/(-6)^2 = 8/36 = 2/9
g''(4) = 8/2^2 = 8/4 = 2
Since g''(-4) > 0 and g''(4) > 0, the graph of g(x) has neither a local maximum nor a local minimum at x = -4 and x = 4.
3. To determine where the graph of g is concave down, we can check the sign of the second derivative.
Since g''(x) > 0 for all x ≠ 2, the graph of g is concave up on the entire domain of the function. There are no intervals where the graph of g is concave down.
4. To find the equation of the tangent line to the graph of g at x = 3, we can use the point-slope form of a linear equation. The slope of the tangent line is the value of the derivative g'(3):
g'(3) = (3^2 – 16)/(3 – 2) = (9 – 16)/(1) = -7
Using the known point (3, g(3)) = (3, 4) and the slope -7, the equation of the tangent line is:
y - 4 = -7(x - 3)
Simplifying, we get:
y = -7x + 25
5. To determine whether the tangent line lies above or below the graph at x = 3, we can compare the y-values of the tangent line and the function at x = 3. Evaluate both the function and the tangent line at x = 3:
g(3) = 4
Substituting x = 3 into the equation of the tangent line:
y = -7(3) + 25 = -21 + 25 = 4
Both the function g and the tangent line have the same y-value at x = 3, which is 4. Therefore, the tangent line lies exactly on top of the graph at this point.
To find the critical values of the function g(x), we need to determine where the derivative is equal to zero or undefined.
1. Set the derivative g'(x) equal to zero:
(x^2 – 16)/(x – 2) = 0
Solve for x:
x^2 – 16 = 0
(x + 4)(x - 4) = 0
x + 4 = 0 -> x = -4
x - 4 = 0 -> x = 4
So, x = -4 and x = 4 are the critical values of g.
2. Determine if the graph of g has local maximum, local minimum, or neither at these critical values.
To do this, we analyze the sign of the derivative around the critical values.
For x < -4, pick a value such as -5:
g'(-5) = ((-5)^2 - 16)/(-5 - 2) = (25 - 16)/(-7) = 9/-7 < 0
For -4 < x < 4, pick a value such as 0:
g'(0) = (0^2 - 16)/(0 - 2) = (-16)/(-2) = 8 > 0
For x > 4, pick a value such as 5:
g'(5) = (5^2 - 16)/(5 - 2) = (25 - 16)/(3) = 9/3 > 0
Since the sign changes from negative to positive, g(x) has a local minimum at x = -4. And since it changes from positive to negative, g(x) has a local maximum at x = 4.
3. To determine where the graph of g is concave down, we need to analyze the concavity of g(x).
The concavity is determined by the sign of the second derivative, which is the derivative of the derivative.
Find the second derivative:
g''(x) = (d/dx) (x^2 – 16)/(x - 2)
= [(2x)(x - 2) - (x^2 - 16)(1)]/(x - 2)^2
= (2x^2 - 4x - x^2 + 16)/(x - 2)^2
= (x^2 - 4x + 16)/(x - 2)^2
For the graph to be concave down, g''(x) < 0.
Thus, solve (x^2 - 4x + 16)/(x - 2)^2 < 0.
(x - 4)^2 > 0 is always positive, so the sign depends on (x^2 - 4x + 16).
The discriminant of the quadratic is negative, indicating no real roots.
Therefore, (x^2 - 4x + 16) > 0 for all x.
So, there are no intervals where the graph of g is concave down.
4. To find the equation for the tangent line to the graph of g at x = 3, we need to find the value of g(3) and the derivative g'(3).
g(3) = 4 (Given)
g'(3) = (3^2 – 16)/(3 - 2) = (9 - 16)/1 = -7
So, the slope of the tangent line is -7. We can use the point-slope form of a linear equation to write the equation of the tangent line:
y - y₁ = m(x - x₁)
y - 4 = -7(x - 3)
y - 4 = -7x + 21
y = -7x + 25
Thus, the equation for the tangent line to the graph of g at the point where x = 3 is y = -7x + 25.
5. To determine if the tangent line lies above or below the graph at x = 3, we compare the value of g(3) with the value of y on the tangent line.
g(3) = 4
Substituting x = 3 into the equation of the tangent line, we get:
y = -7(3) + 25 = -6
The tangent line at x = 3 passes below the graph as the y-value (-6) is less than the function value (4).
Therefore, the tangent line lies below the graph at this point.