The function is:
f ( x ) = x² - x - 72
This is a quatdratic function. So we can find the zeroes of the function with the formula:
x 1/2 = ( - b +/- √(b² - 4 ac) ) / ( 2a )
And we have: a = 1, b = - 1 and c = - 72
x 1/2 = ( 1 +/- √((-1)² - 4 · 1· ( - 72 )) ) / 2
x 1/2 = ( 1 +/- √( 1 + 288 ) ) / 2
x 1/2 = ( 1 +/- √289 ) / 2
x 1/2 = ( 1 +/- 17 ) / 2
x 1 = ( 1 - 17 ) 2 = - 16 / 2 = - 8
x 2 = ( 1 + 17 ) / 2 = 18 / 2 = 9
Answer: The zeroes are - 8 and 9.
The Mean Value Theorem:
If a function is continuous on [ a, b ] and differentiable on ( a , b ) than there is a point c in ( a, b ) such that:
f ` ( c )= ( f ( b ) - f ( a ) ) / ( b - a )
f ` ( c ) = ( f ( 2 ) - f ( 0 ) ) / ( 2 - 0 )
f `( x ) = 10 x - 3
f ` ( c ) = 10 c - 3
2 f ` ( c ) = 16 - 2
f ` ( c ) = 7
7 = 10 c - 3
c = 1
Answer:
Yes, the function is continuous on [ 0, 2 ] and differentiable on ( 0, 2 ).
A rate in which the numerator and the denominator are equal to a rate of one
One way of doing this would be to look at the πr² as if it were a single term. Then we could divide both sides by πr², which leaves h = V/πr².
