The function f is defined by f(x)= (25-x^2)^(1/2) for -5 less than or = x less than or = 5
A) find f'(x)
B) write an equation for the tangent to the graph of f at x=3
C) let g be the function defined by g(x)= (f(x) for -5<or=x<or=-3)and (x+7 for -3<or=x<or=5)
D) find the value of the intagral 5 over 0 of x(25-x^2)^(1/2)dx
A) To find f'(x), we need to differentiate the function f(x). Using the chain rule, we can differentiate (25 - x^2)^(1/2) as follows:
Let u = 25 - x^2, then f(x) = u^(1/2)
By the chain rule, f'(x) = (1/2) * u^(-1/2) * du/dx
Now, let's find du/dx.
du/dx = d/dx (25 - x^2)
= -2x
Substituting these values into f'(x), we have:
f'(x) = (1/2) * (25 - x^2)^(-1/2) * (-2x)
Simplifying, we get:
f'(x) = -x / (25 - x^2)^(1/2)
B) To find the equation of tangent to the graph of f at x = 3, we first need to find the slope of the tangent. The slope of a tangent is equal to the derivative of the function at that point.
So, to find the slope, substitute x = 3 into the expression for f'(x):
f'(3) = -3 / (25 - 3^2)^(1/2)
= -3 / (25 - 9)^(1/2)
= -3 / (16)^(1/2)
= -3 / 4
Now that we have the slope, we can find the equation of the tangent using the point-slope form:
y - y1 = m(x - x1)
We already have the point (x1, y1) = (3, f(3)), but we need to find f(3) first. We'll substitute x = 3 into the function f(x):
f(3) = (25 - 3^2)^(1/2)
= (25 - 9)^(1/2)
= (16)^(1/2)
= 4
So, the point is (x1, y1) = (3, 4), and the slope is m = -3/4.
Using the point-slope form, we get the equation of the tangent:
y - 4 = (-3/4)(x - 3)
C) Let's break down the function g(x) into two separate pieces for each interval: -5 ≤ x ≤ -3 and -3 ≤ x ≤ 5.
For -5 ≤ x ≤ -3:
g(x) = f(x) = (25 - x^2)^(1/2)
For -3 ≤ x ≤ 5:
g(x) = x + 7
D) To find the integral from 0 to 5 of x(25 - x^2)^(1/2), we can use the substitution method.
Let u = 25 - x^2, then du = -2x dx. So we have:
∫[0 to 5] x(25 - x^2)^(1/2) dx
= ∫[0 to 5] -1/2 (25 - x^2)^(1/2) d(25 - x^2)
Substituting u = 25 - x^2, we get:
= -1/2 ∫[25 to 0] u^(1/2) du
Integrating u^(1/2), we get:
= -1/2 * (2/3) u^(3/2) |[25 to 0]
= -1/2 * (2/3) (25 - 0)^(3/2)
= -1/2 * (2/3) (25)^(3/2)
= -1/3 * 25 * 5√25
= -125/3 * 5
= -625/3
To solve these problems, we will go step by step:
A) To find f'(x), we need to differentiate the given function f(x) = (25 - x^2)^(1/2). We can use the power rule for differentiation, which states that if we have a function g(x) = x^n, then g'(x) = nx^(n-1).
Differentiating f(x) = (25 - x^2)^(1/2) using the power rule, we get:
f'(x) = (1/2)(25 - x^2)^(-1/2) * (-2x)
Simplifying further, we have:
f'(x) = -x / (25 - x^2)^(1/2)
B) We need to find the equation of the tangent to the graph of f at x = 3. To do this, we will use the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where (x1, y1) represents a point on the line and m represents the slope.
First, let's find the slope. We can substitute x = 3 into f'(x) to get the slope:
f'(3) = -3 / (25 - 3^2)^(1/2)
Simplifying further, we have:
f'(3) = -3 / (25 - 9)^(1/2)
f'(3) = -3 / 16^(1/2)
f'(3) = -3 / 4
f'(3) = -3/4
Now, let's find the y-coordinate for the point (3, f(3)). We substitute x = 3 into f(x):
f(3) = (25 - 3^2)^(1/2)
f(3) = (25 - 9)^(1/2)
f(3) = 16^(1/2)
f(3) = 4
So, the point on the graph is (3, 4) and the slope is -3/4. Now, we can plug these values into the point-slope form of the equation of a line:
y - 4 = (-3/4)(x - 3)
Simplifying further, we have:
y - 4 = (-3/4)x + (9/4)
y = (-3/4)x + (9/4) + 4
y = (-3/4)x + (9/4) + 16/4
y = (-3/4)x + 25/4
Therefore, the equation of the tangent to the graph of f at x = 3 is y = (-3/4)x + 25/4.
C) To define the function g(x), we need to split it into two parts based on the given conditions.
For -5 ≤ x ≤ -3, g(x) = f(x). Substituting f(x) = (25 - x^2)^(1/2), we have:
g(x) = (25 - x^2)^(1/2)
For -3 < x ≤ 5, g(x) = x + 7.
D) Let's find the value of the integral ∫(5 to 0) x(25 - x^2)^(1/2) dx.
To evaluate this integral, we need to first calculate the antiderivative (or indefinite integral) of the integrand, which is x(25 - x^2)^(1/2).
The antiderivative of x(25 - x^2)^(1/2) can be found by using the substitution method.
Let's substitute u = 25 - x^2:
Then, du/dx = -2x, and dx = -du / (2x).
Now, rewriting the original integral in terms of u:
∫[5 to 0] x(25 - x^2)^(1/2) dx = ∫[5 to 0] x (u^(1/2)) dx
= ∫[u = 25 to u = 0] (-1/2) du
Integrating (u^(1/2)), we get ((2/3)u^(3/2)) as the antiderivative.
Now, evaluating the definite integral:
∫[u = 25 to u = 0] (-1/2) du = [(-1/2) * ((2/3)u^(3/2))] evaluated from 0 to 25
= (-1/2) * ((2/3)25^(3/2) - (2/3)0^(3/2))
= (-1/2) * ((2/3)25^(3/2) - 0)
Simplifying further, we have:
= (-1/2) * ((2/3) * 25 * (25)^(1/2))
= (-1/2) * ((50/3) * (25)^(1/2))
= -25/3 * (25)^(1/2)
= -25/3 * 5
= -125/3
Therefore, the value of the integral ∫(5 to 0) x(25 - x^2)^(1/2) dx is -125/3.