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Verify MVT hypotheses and find the guaranteed value of c

Consider the function f(x)=x2−4x+3f(x) = x^2 - 4x + 3 on the closed interval [1,4][1,4].

The function f(x)=x2−4x+3f(x) = x^2 - 4x + 3 is continuous on [1,4][1,4] and differentiable on (1,4)(1,4), so the Mean Value Theorem guarantees at least one number cc in (1,4)(1,4) such that

f′(c)=f(4)−f(1)4−1.f'(c) = \frac{f(4) - f(1)}{4-1}.

Find the value of cc that satisfies this equation.

Hint

First compute f(4)f(4) and f(1)f(1) to get the average rate of change over [1,4][1,4]. Then compute f′(x)f'(x), set it equal to that average rate of change, and solve for cc, checking that your answer lies in (1,4)(1,4).

Step-by-step solution

First evaluate the function at the endpoints: f(4)=16−16+3=3f(4) = 16 - 16 + 3 = 3 and f(1)=1−4+3=0f(1) = 1 - 4 + 3 = 0. The average rate of change over [1,4][1,4] is f(4)−f(1)4−1=3−03=1\frac{f(4)-f(1)}{4-1} = \frac{3-0}{3} = 1.

Next, differentiate: f′(x)=2x−4f'(x) = 2x - 4. Setting f′(c)=1f'(c) = 1 gives 2c−4=12c - 4 = 1, so c=52c = \frac{5}{2}. Since 52\frac{5}{2} lies in the open interval (1,4)(1,4), it is the valid value guaranteed by the Mean Value Theorem.

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