Question

9–12 ■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers\(c\)that satisfy the conclusion of the Mean Value Theorem.

9.\(f(x) = 2{x^2} - 3x + 1\), \((0,2)\)

Short answer

The numbers \(c = 1\) satisfies the conclusion of Mean Value Theorem.

Step-by-step solution

Given

The function \(f(x) = 2{x^2} - 3x + 1\) is a polynomial.

Concept of Mean Value Theorem

Assume\(f\)to be a function satisfying the following properties.

(1)\(f\)Continuous on the interval\((a,b)\).

(2)\(f\)Is differentiable on the interval\((a,b)\).

Then there is a number\(c\)in\((a,b)\)such that\({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently,\(f(b) - f(a) = {f^\prime }(c)(b - a)\).

Check the conditions of Mean Value Theorem

1. Since the function\(f(x) = 2{x^2} - 3x + 1\)is a polynomial function, it is continuous on\(R\).

Therefore, the function is continu\(f(x)\)ous on the closed interval\((0,2)\).

2. Since the function \(f(x) = 2{x^2} - 3x + 1\) is a polynomial function, it is differentiable everywhere.

Therefore, the function is differentiable on the open interval\((0,2)\).

Hence, the function\(f(x)\)satisfies the conditions of Mean Value Theorem.

Since the number c satisfies the conditions of Mean Value Theorem, it should lie on the open interval \((0,2)\).

Differentiate the function and find the values

Find the derivative of .

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {2{x^2} - 3x + 1} \right)\\ &= 2(2x) - 3(1) + 0\\ &= 4x - 3\end{aligned}\)

Replace\({\rm{x }} \to c\)in\({f^\prime }(x)\).

\({f^\prime }(c) = 4c - 3\)

Obtain the functional values at the end points of the given interval.

Substitute\(0{\rm{ }} \to x\)in\(f(x)\).

\(\begin{aligned}{c}f(a) &= f(0)\\ &= 2{(0)^2} - 3(0) + 1\\ &= 1\end{aligned}\)

Substitute\(2{\rm{ }} \to x\)in\(f(x)\).

\(\begin{aligned}{c}f(b) &= f(2)\\ &= 2{(2)^2} - 3(2) + 1\\ &= 8 - 6 + 1\\ &= 3\end{aligned}\)