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
There's nothing to figure out for C. They're just defining a new function. Not sure why it's even there.
For D, it's pretty straightforward. Let
u = 25-x^2
then du = -2x dx and we have
∫[0,5] x√(25-x^2) dx
= ∫[5,0] -1/2 √u du
and it's clear sailing from there on
A: d/dx √(25-x^2) = -x/√(25-x^2)
B. At x=3, f(x)=4, so
f'(3) = -3/4
Now you have a point (3,4) and a slope: -3/4
y-4 = -3/4 (x-3)
i found A and B, just checking if they're right. C i can never remember how to do these. D i haven't tried yet but i could probably figure out but i figured since i was putting the rest up i would do it as well
thank youu
A) To find f'(x), we need to find the derivative of the function f(x) = (25 - x^2)^(1/2). We can do this by using the chain rule.
First, we rewrite the function as f(x) = (25 - x^2)^(1/2) = (25 - x^2)^0.5.
Next, we differentiate the function with respect to x, using the chain rule:
f'(x) = 0.5(25 - x^2)^(-0.5)(-2x).
Therefore, f'(x) = -x / (25 - x^2)^(0.5).
B) To find the equation of the tangent at x = 3, we need both the slope and a point on the tangent line. The slope is given by f'(x), evaluated at x = 3.
To find the slope, substitute x = 3 into the expression for f'(x):
f'(3) = -(3) / (25 - (3)^2)^(0.5) = -3 / 4.
Now, we need a point on the graph. We can find the corresponding y-value by substituting x = 3 into the original function f(x):
f(3) = (25 - (3)^2)^(0.5) = (25 - 9)^(0.5) = 4.
So, the coordinates of the point on the graph are (3, 4).
Using the point-slope form of a straight line:
y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values:
y - 4 = (-3/4)(x - 3).
Simplifying:
y - 4 = (-3/4)x + 9/4.
Therefore, the equation of the tangent to the graph of f at x = 3 is y = (-3/4)x + 9/4.
C) The function g is defined piecewise. For -5 ≤ x ≤ -3, g(x) = f(x), and for -3 < x ≤ 5, g(x) = x + 7.
So, for -5 ≤ x ≤ -3, g(x) = f(x) = (25 - x^2)^(0.5).
And for -3 < x ≤ 5, g(x) = x + 7.
D) To find the value of the integral of x(25 - x^2)^(0.5) dx from 0 to 5, we need to find the antiderivative of the integrand and evaluate it at the upper and lower limits.
The integrand x(25 - x^2)^(0.5) can be simplified using the substitution u = 25 - x^2:
Let u = 25 - x^2.
Then du = -2x dx, or dx = (-du / (2x)).
Substituting these values into the integrand, we have:
∫ x(25 - x^2)^(0.5) dx = ∫ (-u^(0.5)) du / 2.
Now, we integrate with respect to u:
∫ (-u^(0.5)) du / 2 = -∫ (u^(0.5)) du / 2.
Using the power rule of integration: (u^(n+1)) / (n+1), where n ≠ -1, we have:
-∫ (u^(0.5)) du / 2 = -((u^(0.5 + 1)) / (0.5 + 1)) + C.
Simplifying and substituting back u = 25 - x^2:
-((u^(0.5 + 1)) / (0.5 + 1)) + C = -(2/3)(25 - x^2)^(1.5) + C.
Now, we evaluate the antiderivative at the upper and lower limits:
∫[0 to 5] x(25 - x^2)^(0.5) dx = -((2/3)(25 - 5^2)^(1.5)) + (2/3)(25 - 0^2)^(1.5).
Simplifying:
-((2/3)(25 - 25)^(1.5)) + (2/3)(25)^(1.5) = 2/3 * 25^(1.5).
Therefore, the value of the integral from 0 to 5 is 2/3 * 25^(1.5).
do you have no idea at all about any of these questions? Hard to believe this would come up till the topics have been covered. They look like pretty straightforward calculus and algebra operations.
How far do you get?