AboutSteve Holleran Expertise I can help with all math questions from basic math to Calculus.
Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.
Experience 33 years teaching experience in NJ public schools
Education/Credentials B.S. Mathematics : Wake Forest University 1972
M.S. Mathematics : Monmouth University 1981
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:
