integration and derivative problems

[Hockeyman]

1) if f(x)= x^3 -x + 3 and if c is the only real number such that f(c)=0 then c is between
(a) -2 and -1 (b) -1 and 0 (c) 0 and 1 (d) 1 and 2 (e) 2 and 3

2) f(x)= x^2 +2 x< or =1
f(x)= 2x + 1 x> 1

then f'(1) is


3) for what valure of k will 8x + k/x^2 have a relative maximum at x=4?

4) the area of the region enclosed by the y axis and the graphs of y=3cosx and y=x is

5) if f is a function such that f(0)=1, f(1)=2, and f(n)= f(n-2) / f(n-1) for all integers n> or = 0, what is the value of f(4)?

 

[o_O]

What are your thoughts on these questions?

 

[skeeter]

1) if f(x)= x^3 -x + 3 and if c is the only real number such that f(c)=0 then c is between
(a) -2 and -1 (b) -1 and 0 (c) 0 and 1 (d) 1 and 2 (e) 2 and 3

look up the Intermediate Value Theorem

2) f(x)= x^2 +2 x< or =1
f(x)= 2x + 1 x> 1

then f'(1) is

first check if f(x) is continuous at x = 1... if so, then check if the derivative f'(x) is also continuous at x = 1 ... you know the requirements for a function to be continuous at a point, don't you?


3) for what valure of k will 8x + k/x^2 have a relative maximum at x=4?

relative maxima of a function occur at critical values where the function is defined and the function's derivative changes sign from positive to negative

4) the area of the region enclosed by the y axis and the graphs of y=3cosx and y=x is

the area between two functions is \(\displaystyle \int_a^b f(x) - g(x) \, dx\) where f(x) > g(x).

5) if f is a function such that f(0)=1, f(1)=2, and f(n)= f(n-2) / f(n-1) for all integers n> or = 0, what is the value of f(4)?

start by figuring out the value for f(2) ... then f(3) ... then ?

 

[Hockeyman]

2) a function is continuos when the lim f(x) exists, f(a) exists, and the lim f(x) = f(a), so i found that f(1)=3 for both equations and the lim as x-a=3, then i found that f'(1)= 2, is this correct?

3) so therefore i would have to find the derivative of that function and set it equal to zero, then put 4 in for x and solve for k, correct?

4) using my calculator i found the bounds to be [0,1.17], and when i calculated the value it came out to be 2.076

5) i found f(2)= 1/2, then f(3)= 4, then using these values i found that f(4)= 1/8

 

[skeeter]

2) a function is continuos when the lim f(x) exists, f(a) exists, and the lim f(x) = f(a), so i found that f(1)=3 for both equations and the lim as x-a=3, then i found that f'(1)= 2, is this correct?

f'(1) = 2 ... correct

3) so therefore i would have to find the derivative of that function and set it equal to zero, then put 4 in for x and solve for k, correct?

little more to it than that ... you'll also have to check if there is a maximum at x = 4. just because the derivative equals zero doesn't guarantee that the function has a maximum there. additional hint ... if a maximum exists at x = 4, what should the sign of the 2nd derivative be at x = 4?

4) using my calculator i found the bounds to be [0,1.17], and when i calculated the value it came out to be 2.076

that would be the area of the region in quadrant I ... do you see another "piece" of the defined region in quadrant III?

5) i found f(2)= 1/2, then f(3)= 4, then using these values i found that f(4)= 1/8
fine