Consider a differentiable functionf having domain all positive real numbers, and for which it is known that f'(x)=(4-x)^-3 for x.0
a. find the x-coordinate of the critical point f. DEtermine whether the point is a relative maximum or minomum, or neither for the fxn f. justify your answer.
b) find all intervals on which the graph of f is concave down. jusity answer.
c. given that f(1)=2 determine the fxn f.
i got a but how do you do b and c?
c)
if f'(x) = (4-x)^-3
then f(x) = (1/2)(4-x)^-2 + c
given f(1) = 2
2 = (1/2)(4-1)^-2 + c
2 = (1/2)(1/9) + c
c = 2 - 1/18 = 35/18
f(x) = (1/2)(4-x)^-2 + 35/18
f''(x) = -3(4-x)^-4 (-1) = 3/(4-x)^4
the graph is concave up when f''(x) > 0
the graph is concave down when f''(x) < 0
since (4-x)^4 is always positive, for x≠4
then 3/(4-x)^4 is always positive.
So the curve is concave up for all x's in the domain.
here is what Wolfram thinks of your function
http://www.wolframalpha.com/input/?i=%281%2F2%29%284-x%29%5E-2+%2B+35%2F18
To find the intervals on which the graph of f is concave down and to determine the function f given f(1) = 2, you need to use the second derivative of f, denoted as f''(x).
b) To determine the intervals on which the graph of f is concave down, you need to find where the second derivative f''(x) is negative. Here's how you can do it:
1. Find the second derivative:
f'(x) = (4 - x)^-3
To find f''(x), take the derivative of f'(x):
f''(x) = d/dx [(4 - x)^-3]
2. Simplify the second derivative:
To simplify, we can rewrite the expression of f'(x) as:
f'(x) = (4 - x)^(-3) = (4 - x)^(-3*1) = (4 - x)^-3
Now, differentiate using the power rule:
f''(x) = 3(4 - x)^(-3 - 1) * (-1) = -3(4 - x)^-4
3. Determine when f''(x) is negative:
Since f''(x) = -3(4 - x)^-4, we need to find when f''(x) is less than zero (-3 < 0).
-3(4 - x)^-4 < 0
Multiplying both sides by (-1), we get:
3(4 - x)^-4 > 0
Since 3 is positive, the sign of (4 - x)^-4 determines whether f''(x) is positive or negative.
When (4 - x)^-4 > 0:
-4 is an even power, so (4 - x)^-4 > 0 for all values of x except at x = 4.
Therefore, the graph of f is concave down for all values of x except x = 4.
c) To find the function f given f(1) = 2, we can use the information about f'(x) and integrate it. Here's how:
1. Integrate f'(x) to find f(x):
We know that f'(x) = (4 - x)^-3.
To find f(x), integrate f'(x):
∫ f'(x) dx = ∫ (4 - x)^-3 dx
2. Apply the power rule for integrals:
∫ (4 - x)^-3 dx = -1/(2 - x)^2 + C
Where C is the constant of integration.
3. Determine the value of C using f(1) = 2:
Since f(1) = 2, we can substitute the value into the equation:
-1/(2 - 1)^2 + C = 2
Simplifying, we get:
-1/1^2 + C = 2
-1 + C = 2
C = 2 + 1
C = 3
4. Final function f(x):
Substituting the value of C back into the equation, we get:
f(x) = -1/(2 - x)^2 + 3
Therefore, the function f(x) is f(x) = -1/(2 - x)^2 + 3, and the intervals on which the graph of f is concave down are all values of x except x = 4.