Posted by Aparna on Tuesday, March 19, 2013 at 3:05pm.
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
- Calc - Steve, Tuesday, March 19, 2013 at 3:07pm
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?
- Calc - Aparna, Tuesday, March 19, 2013 at 3:12pm
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
- Calc - Steve, Tuesday, March 19, 2013 at 3:19pm
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
- Calc - Aparna, Tuesday, March 19, 2013 at 3:21pm
- Calc - Steve, Tuesday, March 19, 2013 at 3:22pm
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)
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