Graph the function f(x) = x + 5/x in the viewing rectangle [0, 12] by [0, 12]. (Do this on paper. Your instructor may ask you to turn in this graph.)
Then graph the secant line that passes through the points (1, 6) and (10, 10.5). (Do this on paper. Your instructor may ask you to turn in this graph.)
Find the number c that satisfies the conclusion of the Mean Value Theorem for this function f and the interval [1, 10].
c =
10.In the viewing rectangle [-4, 4] by [-20, 20], graph the function f(x) = x3 – 3x and its secant line through the points (-3, -18) and (3, 18). (Do this on paper. Your instructor may ask you to turn in this graph.)
Find the values of the numbers c that satisfy the conclusion of the Mean Value Theorem for the interval [-3, 3].
c = (smaller value)
c = (larger value)
The Correct Answer and Explanation is :
The number ( c ) that satisfies the conclusion of the Mean Value Theorem for the function ( f(x) = x + \frac{5}{x} ) on the interval ([1,10]) is:
[
c = 3.162
]
For the function ( g(x) = x^3 – 3x ) on the interval ([-3,3]), the values of ( c ) that satisfy the Mean Value Theorem are:
[
c = -1.732, \quad c = 1.732
]
Explanation (300 Words):
The Mean Value Theorem (MVT) states that for a function ( f(x) ) that is continuous on the closed interval ([a, b]) and differentiable on the open interval ( (a, b) ), there exists at least one number ( c ) in ( (a, b) ) such that:
[
f'(c) = \frac{f(b) – f(a)}{b – a}
]
First Function: ( f(x) = x + \frac{5}{x} )
- We first compute the average rate of change (secant slope) from ( x = 1 ) to ( x = 10 ):
[
\frac{f(10) – f(1)}{10 – 1} = \frac{10 + 0.5 – (1 + 5)}{9} = \frac{10.5 – 6}{9} = \frac{4.5}{9} = 0.5
]
- The derivative of ( f(x) ) is:
[
f'(x) = 1 – \frac{5}{x^2}
]
- Solving ( f'(c) = 0.5 ) gives ( c = 3.162 ), which is in ( (1,10) ).
Second Function: ( g(x) = x^3 – 3x )
- The secant slope for ( g(x) ) from ( x = -3 ) to ( x = 3 ):
[
\frac{g(3) – g(-3)}{3 – (-3)} = \frac{18 – (-18)}{6} = \frac{36}{6} = 6
]
- The derivative of ( g(x) ) is:
[
g'(x) = 3x^2 – 3
]
- Solving ( g'(c) = 6 ) gives ( c = \pm 1.732 ), both in ( (-3,3) ).
Thus, these values satisfy MVT.