Consider a differentiable function f having domain all positive real numbers, and for which it is known that f'(x)=(4-x)x^-3 for x>0.
A. Find the x-coordinate of the critical point of f. Determine whether the point is a relative maximum, a relative minimum, or neither for the function f. Justify your answer.
B. Find all intervals in which the graph of f is concave down. Justify your answer.
C. Given that f(1)=2, determine the function of f.
A- I know for max and min you use first derivative and equal the function to 0 for critical numbers.
B-I know concave down is determined from the second derivative.
C-Do not have any idea how to do it
Thanks for your help
http://apcentral.collegeboard.com/apc/public/repository/ap11_calc_ab_form_b_q4.pdf
A. To find the critical point of the function f, we need to find the values of x for which f'(x) = 0 or f'(x) does not exist. Let's set f'(x) = 0 and solve for x:
4 - x = 0
x = 4
So the critical point of f is at x = 4. To determine whether it is a relative maximum, relative minimum, or neither, we can analyze the sign of f'(x) around x = 4.
When x < 4, f'(x) = (4 - x)x^(-3) < 0.
When x > 4, f'(x) = (4 - x)x^(-3) > 0.
This means that the function is decreasing for x < 4 and increasing for x > 4. Therefore, at x = 4, the function f has a relative minimum.
B. To find the intervals in which the graph of f is concave down, we need to determine the sign of the second derivative, f''(x). Let's differentiate f'(x):
f''(x) = d/dx [(4 - x)x^(-3)]
= 3x^(-4) - 4x^(-5)
For the graph to be concave down, f''(x) should be negative. Let's solve f''(x) < 0:
3x^(-4) - 4x^(-5) < 0
Since the function is defined only for positive real numbers, we can multiply both sides of the inequality by x^5 (which is positive):
3x - 4 < 0
3x < 4
x < 4/3
Therefore, the graph of f is concave down for x < 4/3.
C. To determine the function f(x), we need to integrate f'(x). Let's integrate (4 - x)x^(-3):
∫[(4 - x)x^(-3)] dx = ∫[4x^(-3) - x^(-2)] dx
= [-2x^(-2) + x^(-1)] + C
= -2/x^2 + 1/x + C
Given that f(1) = 2, we can solve for C:
-2/1^2 + 1/1 + C = 2
-2 + 1 + C = 2
C = 3
So the function f(x) is:
f(x) = -2/x^2 + 1/x + 3
A. To find the critical point, we need to solve for x when the derivative f'(x) equals zero or does not exist. In this case, the derivative f'(x) is given as (4 - x)x^-3.
Setting f'(x) equal to zero, we have:
(4 - x)x^-3 = 0
Since x^-3 cannot be zero, we can ignore that part of the equation. Therefore, we are left with:
4 - x = 0
Solving for x, we find:
x = 4
To determine whether this critical point is a relative maximum, relative minimum, or neither, we can further analyze the behavior of the function.
We already have the first derivative, f'(x), which is always positive for positive x. This tells us that the function is monotonically increasing for x > 0.
Since the function is increasing, we know that the point (4, f(4)) corresponds to a relative minimum of the function f.
B. To find the intervals in which the graph of f is concave down, we need to analyze the second derivative f''(x). The second derivative can be found by taking the derivative of the first derivative.
First, let's find f''(x):
f''(x) = d/dx(4 - x)x^-3
Using the product rule and the power rule, we can differentiate the above expression as follows:
f''(x) = (-1)x^-3 + (4 - x)(-3)x^-4
Simplifying, we get:
f''(x) = -x^-3 - 12x^-4 + 3x^-3
To determine when the function is concave down, we need to find where the second derivative is negative.
Setting f''(x) less than zero, we have:
-x^-3 - 12x^-4 + 3x^-3 < 0
Multiplying through by -x^4 to eliminate the negative signs, we have:
x + 12 - 3x < 0
Combining like terms, we get:
-2x + 12 < 0
Solving for x, we find:
x > 6
Therefore, the graph of f is concave down for x > 6.
C. To determine the function f, we need to integrate the derivative f'(x) with respect to x.
Starting with:
f'(x) = (4 - x)x^-3
Integrating both sides with respect to x, we get:
f(x) = ∫ (4 - x)x^-3 dx
To integrate the right hand side, we can rewrite the integral as:
f(x) = ∫ (4 - x)(1/x^3) dx
Expanding the integrand, we get:
f(x) = ∫ (4/x^3 - x/x^3) dx
Simplifying further, we have:
f(x) = 4∫ (1/x^3) dx - ∫ (x/x^3) dx
Integrating each term separately, we get:
f(x) = 4(-1/2x^2) - (1/2x^2) + C
Combining like terms, we have:
f(x) = -2/x^2 + C
Given that f(1) = 2, we can substitute x = 1 into the function f(x) to find the value of the constant C:
2 = -2/1^2 + C
2 = -2 + C
C = 4
Therefore, the function f is:
f(x) = -2/x^2 + 4
A. To find the critical point of f, we need to find the values of x for which f'(x) = 0 or f'(x) is undefined. In this case, f'(x) is defined for all positive real numbers, so we set f'(x) = 0 and solve for x:
(4 - x)x^(-3) = 0
Since x^(-3) cannot be zero, we can multiply both sides by x^3 to get:
4 - x = 0
Solving for x, we find:
x = 4
So the x-coordinate of the critical point of f is 4.
To determine whether the critical point is a relative maximum, minimum, or neither, we can use the second derivative test. Since we haven't found the second derivative yet, we'll move on to part B and come back to this question later.
B. To determine the intervals in which the graph of f is concave down, we need to find the intervals where f''(x) < 0. To find f'', we differentiate f' with respect to x:
f'(x) = (4 - x)x^(-3)
Using the product rule and the power rule for differentiation, we get:
f''(x) = -3(4 - x)x^(-4) + (-1)x^(-3)
Simplifying, we have:
f''(x) = 3x^(-4) - 4x^(-3)
To find the intervals where f''(x) < 0, we can examine the sign changes of f''(x).
At x > 0, both terms in f''(x) are positive, so f''(x) > 0. Therefore, there are no intervals where the graph of f is concave down.
Now let's go back to part A and use the second derivative test to determine whether the critical point (x = 4) is a relative maximum, minimum, or neither.
To apply the second derivative test, we evaluate f''(x) at the critical point.
f''(4) = 3(4)^(-4) - 4(4)^(-3)
Simplifying, we have:
f''(4) = 3/256 - 4/64
f''(4) = 3/256 - 1/16
f''(4) = 3/256 - 16/256
f''(4) = -13/256
Since f''(4) < 0, the second derivative test tells us that the critical point x = 4 is a relative maximum for the function f.
C. To determine the function f, we need to integrate f'(x) with respect to x.
Let's integrate f'(x) = (4 - x)x^(-3) with respect to x:
f(x) = ∫[(4 - x)x^(-3)] dx
Using the power rule for integration, we have:
f(x) = ∫[4x^(-3) - x^(-2)] dx
Integrating each term separately, we get:
f(x) = -2x^(-2) + x^(-1) + C
where C is the constant of integration.
Given that f(1) = 2, we can substitute x = 1 and f(x) = 2 into the equation and solve for C:
2 = -2(1)^(-2) + (1)^(-1) + C
2 = -2 + 1 + C
2 = -1 + C
C = 3
Thus, the function f(x) is:
f(x) = -2x^(-2) + x^(-1) + 3