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.
here's my work so far
(4-x)^-3=0
4x^-3-((x^2)^-3)
4x^-3-x^-6=0
4-x=0.......x^-3
+4..+4......x=0
-x=4
x=-4
if f' = (4-x)^-3
f'' = (-3)(4-x)^-4 (-1) = 3(4-x)^-4
f' and f'' are never zero.
There is a vertical asymptote ate x = 4.
How did you get from line 1 to line 2 of your solution?
f(x) = 1/2 (4-x)^-2 + C
2 = 1/18 + C
umm..i don't really know i i didn't know how to do it.. i don't even think i need that part anyway.....and i checked my work and it was x=4 not -4
how did you find f(x)? i didn't know how to do it??
To find the x-coordinate of the critical point, we need to find the values of x for which f'(x) is equal to zero or undefined. In this case, f'(x) = (4-x)^-3.
To determine when f'(x) = 0, we can set (4-x)^-3 = 0. However, the expression (4-x)^-3 cannot equal zero because a number raised to a negative power cannot be zero. So, there are no values of x for which f'(x) = 0.
Next, let's consider when f'(x) is undefined. This occurs when the denominator of the expression (4-x)^-3 is equal to zero. So, (4-x) should equal zero:
4-x = 0
x = 4
Therefore, the x-coordinate of the critical point f is x = 4.
To determine whether this critical point is a relative maximum, minimum, or neither, we can use the second derivative test. However, we still need to find the second derivative of f(x):
f'(x) = (4-x)^-3
Using the power rule to find the derivative:
f''(x) = 3(4-x)^-4 * (-1)
Simplifying:
f''(x) = -3/(4-x)^4
Now, we can evaluate f''(4) to determine if the critical point is a maximum, minimum, or neither:
f''(4) = -3/(4-4)^4
f''(4) = -3/0
Since the second derivative is undefined, the second derivative test is inconclusive, and we cannot determine whether the critical point is a relative maximum, minimum, or neither.
Moving on to part b, to find the intervals on which the graph of f is concave down, we need to determine where the second derivative, f''(x), is negative.
f''(x) = -3/(4-x)^4
Since the denominator is squared, it is always positive. Therefore, f''(x) will be negative if the numerator, -3, is negative.
So, f''(x) < 0 when -3 < 0, which is always true.
Hence, the graph of f is concave down for all positive real numbers.
Lastly, part c asks us to find the function f(x) given that f(1) = 2.
To find the function, we will integrate the derivative f'(x) = (4-x)^-3 with respect to x to get f(x):
∫[(4-x)^-3]dx
Let's use integration by substitution:
Let u = 4-x, then du = -dx
The integral becomes:
-∫u^-3 du
Using the power rule, we can integrate:
-∫u^-3 du = -((-2)/u^2) + C
Substituting u back:
-((-2)/(4-x)^2) + C
Since f(1) = 2, we can substitute x = 1 and solve for C:
-((-2)/(4-1)^2) + C = 2
-((-2)/9) + C = 2
Simplifying and solving for C:
C - (2/9) = 2
C = 2 + (2/9)
C = 20/9
Therefore, the function f(x) is:
f(x) = -((-2)/(4-x)^2) + 20/9