Let f be the function shown below, with domain the closed interval [0, 6]. Let h(x) = integral from 0 to 2x-1 (f(t))dt.
The graph is a semicircle above the x-axis with r=2 and center (2,0) connected to a semicircle below the x-axis with r=1 and center (5,0).
A. Find h'(x)
B. At what x is h(x) a maximum?
So I determined the domain for h(x) = .5 ≤ x ≤ 1.5 and differentiated h(x) to get h'(x) = 2f (2x-1), but I don't know how to solve the rest. Any help is appreciated :)
You sure on that domain?
2x-1 = 0 ==> x = 1/2
2x-1 = 6 ==> x = 7/2
since h'(x) = 2f(2x-1), h(x) is a max when f(2x-1) = 0
f(4) = 0, so 2x-1=4 ==> x = 5/2
thank you! sorry the domain was a mistype.
how do i solve for h'(5/2)?
does the equation integral from 0 to x (2-t) / √4-(t-2)^2 work?
^ the equation for f
nvm i figured it out. thanks for your help !!
To find the derivative of h(x) (h'(x)), we need to apply the Fundamental Theorem of Calculus and the Chain Rule.
A. First, let's rewrite h(x) using the definite integral notation:
h(x) = ∫[0 to 2x-1] f(t) dt
Now, we can differentiate h(x) using the Fundamental Theorem of Calculus and the Chain Rule. The Fundamental Theorem of Calculus states that if a function F(x) is the integral of f(x), then the derivative of F(x) is f(x).
The Chain Rule states that if we have a composition of functions, such as f(g(x)), the derivative of this composition is the derivative of the outer function multiplied by the derivative of the inner function.
So, applying the Chain Rule, we get:
h'(x) = d/dx [∫[0 to 2x-1] f(t) dt]
= f(2x-1) * d/dx (2x-1)
Since f(t) is defined piecewise, we need to consider two cases based on the value of (2x-1):
Case 1: 0 ≤ 2x-1 < 2
In this case, (2x-1) lies within the interval [0,2]. The equation of the semicircle with radius 2 and center (2,0) can be written as:
f(t) = √(4 - (t-2)^2)
So, f(2x-1) = √(4 - ((2x-1)-2)^2) = √(4 - (2x-2)^2) = √(4 - 4(x-1)^2)
Taking the derivative of (2x-1) with respect to x, we get:
d/dx (2x-1) = 2
Case 2: 2 ≤ 2x-1 ≤ 6
In this case, (2x-1) lies within the interval [2,6]. The equation of the semicircle with radius 1 and center (5,0) can be written as:
f(t) = -√(1 - (t-5)^2)
So, f(2x-1) = -√(1 - ((2x-1)-5)^2) = -√(1 - (2x-6)^2) = -√(1 - 4(x-3)^2)
Taking the derivative of (2x-1) with respect to x, we get:
d/dx (2x-1) = 2
Therefore, combining both cases, we can write h'(x) as:
h'(x) = 2f(2x-1)
B. To find at what value of x h(x) is a maximum, we need to find the critical points of h(x). In other words, we need to find the values of x where h'(x) = 0 or h'(x) is undefined.
From our derivative h'(x) = 2f(2x-1), we can see that h'(x) is only zero if f(2x-1) = 0, as the constant factor 2 cannot make the entire expression zero. Since f(t) is defined piecewise, let's consider the two cases for f(t):
Case 1: 0 ≤ 2x-1 < 2
In this case, we need to find the values of x that make √(4 - 4(x-1)^2) = 0. Solving this equation, we get:
4 - 4(x-1)^2 = 0
(x-1)^2 = 1
x-1 = ±1
x = 2 or x = 0
Case 2: 2 ≤ 2x-1 ≤ 6
In this case, we need to find the values of x that make -√(1 - 4(x-3)^2) = 0. However, this is impossible since the square root of a positive quantity will never be zero.
Therefore, the only critical point occurs when x = 2.
In summary, the derivative of h(x) is h'(x) = 2f(2x-1), and the only critical point where h(x) is a maximum is when x = 2.