Expert: Steve Holleran - 11/2/2008

Question
Find the maximum and minimum values of 4sinx + 9/(1+ sinx) for 0<= x<= pi. I wasnt sure if you integrate it as a whole or integrate it into two parts, and how you would do so. Thank you.


Express f(x) = (3x^2 +1)^2/ (x^2) in the form Ax^2 + Bx + C/x^2. Hence evaluate the integral f(x) dx between 2 and 1.

thank you for your help!!

Answer
Hi Eliza,

for the first one, I don't think you want to integrate at all.  
You should apply the Closed Interval Test for extrema:

On a closed interval, a continuous function could have max / min values at:  the endpoints, at points where f' = 0, or at points where f' does not exist.

So,  f(x) = 4sin x + [9/(1+sin x)] on [0, pi]

    f'(x) = 4 cos x - [9cos x /(1 + sin x)^2] = 0

    f'(x) = 0 at 4 cos x = 9 cos x / (1+ sin x)^2

or                4 = 9 / (1 + sin x)^2

or               (1 + sin x)^2 = 9/4

                 1 + sin x = +/- 3/2

                     sin x = 3/2 - 1   or -3/2 - 1

                     sin x = 1/2       or sin x = -5/2 (undefined)

Now sin x = 1/2 at x = pi/6 and 5pi/6 in the mentioned interval.

Also, f'(x) does not exist when 1 + sin x = 0 , meaning sin x = -1
which occurs at -pi/2, but this is outside the interval, so we won't use it.

so, for the closed interval test we have:

f(0) = 9       f(pi/6) = 8    f(5pi/6) = 8   and f(pi) = 9

So the max value is 9 and the min value is 8.


For the second one, expand the square and divide each resulting term by x^2:

(3x^2 + 1)^2 = 9x^4 + 6x^2 + 1 / x^2 = 9x^2 + 6 + 1/x^2

             = 9x^2 + 6 + x^(-2)

so INT(1 to 2) [9x^2 + 6 + x^(-2)] = 3x^3 + 6x - (1/x) evaluated
x = 1 to x = 2 gives  (24 + 12 - 1/2) - (3 + 6 - 1) = 27.5.


Hope this is what you needed.
Steve