Question

The graph of a function \(g\) is shown.

(a) Verify that \(g\) satisfies the hypotheses of the Mean ValueTheorem on the interval \(\left( {0,8} \right)\).

(b) Estimate the value(s) of \(c\) that satisfy the conclusion ofthe Mean Value Theorem on the interval \(\left( {0,8} \right)\).

(c) Estimate the value(s) of \(c\) that satisfy the conclusion ofthe Mean Value Theorem on the interval \(\left( {2,6} \right)\).

Short answer

(a) \(g\) satisfies the hypothesis of the Mean Value Theorem.

(b) The values of \(c\) are \(2.2\) and \(6.4\).

(c) The values of \(c\) are \(3.7\) and \(5.5\).

Step-by-step solution

Mean Value Theorem

Mean Value Theorem states that if a function \(f\) satisfies the following conditions:

1. Continuous on an interval \(\left( {a,b} \right)\).

2. Differentiable on \(\left( {a,b} \right)\).

Then there exists at least one point \(c \in \left( {a,b} \right)\) such as that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\).

(a) Step 2: Verification of Mean Value Theorem

From the graph, we can say that on the interval\(\left( {0,8} \right)\), the function\(g\)is continuous on\(\left( {0,8} \right)\), differentiable on\(\left( {0,8} \right)\).

Thus the function \(g\) satisfies the hypothesis of the Mean Value Theorem.

(b) Step  3: Finding the values of \(c\)

On the interval \(\left( {0,8} \right)\) applying MVT to the function, \(g\) we get;

\(\begin{aligned}{c}g'\left( c \right) &= \frac{{g\left( 8 \right) - g\left( 0 \right)}}{8}\\ &= \frac{{4 - 1}}{8}\\ &= \frac{3}{4}\end{aligned}\)

Now \(g'\left( c \right) = \frac{3}{8}\) when \(c = 2.2\) and \(c = 6.4\).

(c) Step 4: Finding the values of \(c\)

On the interval\(\left( {2,6} \right)\)applying MVT to the function, we get;

\(\begin{aligned}{c}g'\left( c \right) &= \frac{{g\left( 6 \right) - g\left( 2 \right)}}{4}\\ &= \frac{{1 - 3}}{4}\\ &= \frac{{ - 1}}{2}\end{aligned}\)

Now \(g'\left( c \right) = - \frac{1}{2}\) when \(c = 3.7\) and \(c = 5.5\).