The Mean Value Theorem

Question

Consider the function f(x) = 1/x on the interval [3, 11]

(A) Find the average or mean slope of the function on this interval.

Average Slope = ________

(B) By the Mean Value Theorem, we know there exists a c in the open interval (3, 11) such that f '(c) is equal to this mean slope. Find all values of c that work and list them (separated by commas) in the box below.

List the Values : _____________

Answer 1

The Mean Value Theorem

Consider the function f(x) = 1/x on the interval [3, 11]

(A) Find the average or mean slope of the function on this interval. (m_sec - slope of secant)

Average Slope = m_sec (slope of secant)

(B) By the Mean Value Theorem, we know there exists a c in the open interval (3, 11) such that f '(c) is equal to this mean slope. Find all values of c that work and list them (separated by commas) in the box below.

List the Values: _____________

Detailed Solution:

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1) f(x) = 1/x is continuous for all value of x; except x = 0

==> f(x) is continuous on [3, 11]

2) f ^' (x) = -1/x²; f(x) is differentiable for all value of x; except x = 0

==> f(x) is differentiable on ([3, 11])

(A) Average Slope = m_sec (slope of secant) = (f(b) - f(a))/(b - a) = (f(11) - f(3))/(11 - 3) = ((1/11) - (1/3))/(11 - 3) = ((3 - 11)/(33)/(8)) = ((-8)/(33)/(8)) = -1/33

Average Slope = m_sec (slope of secant) = -1/33

(B) By the Mean Value Theorem:

f ^' (c) = (f(b) - f(a))/(b - a) <== m_tan (slope of tangent) = m_sec (slope of secant)

f ^' (c) = (f(11) - f(3))/(11 - 3) ==> -1/c² = -1/33 ==> c² = 33 ==> c = ±√33

c = √33 = 5.74456 <== in the open interval (3, 11) such that f '(c) is equal to this mean slope

Note:

1) m_sec (slope of secant) = -1/33 = Average Slope

2) f ^' (x) = -1/x² = m_tan (slope of tangent)

==> f ^' (c) = -1/c² ==> f ^' (√33) = -1/(√33)² = -1/33

At c = √33, m_sec = m_tan = -1/33