Function Analysis: Understanding the Problem

The question asks us to find a specific point \(x\)   within the interval  \([1, 3]\)   for the function  \(f(x) = x^2 - 2x + 2\). At this point  \(x\), the tangent line to the curve of  \(f(x)\)   must be parallel to the straight line that connects the points  \(f(1)\)   and  \(f(3)\)   on the function's graph. This scenario is a direct application of the Mean Value Theorem (MVT).

The Mean Value Theorem states that if a function  \(f(x)\)   is:

• Continuous on the closed interval  \([a, b]\), and
• Differentiable on the open interval  \((a, b)\),

then there exists at least one point  \(c\)   in  \((a, b)\)   such that the instantaneous rate of change (slope of the tangent at  \(c\)) is equal to the average rate of change over the interval (slope of the secant line connecting  \((a, f(a))\)   and  \((b, f(b))\)). Mathematically, this is expressed as:

\[f'(c) = \frac{f(b) - f(a)}{b - a}\]

Continuity and Differentiability of f(x)

First, let's verify if our given function  \(f(x) = x^2 - 2x + 2\)   satisfies the conditions of the Mean Value Theorem over the interval  \([1, 3]\):

• Continuity:  \(f(x)\)   is a polynomial function. Polynomials are continuous everywhere, so  \(f(x)\)   is continuous on the closed interval  \([1, 3]\).
• Differentiability:  \(f(x)\)   is also differentiable everywhere. Its derivative,  \(f'(x) = 2x - 2\), exists for all  \(x\). Therefore,  \(f(x)\)   is differentiable on the open interval  \((1, 3)\).

Since both conditions are met, the Mean Value Theorem can be applied.

Calculating the Slope of the Secant Line

The problem states that the tangent must be parallel to the straight line joining  \(f(1)\)   and  \(f(3)\). This straight line is called the secant line. We need to find the coordinates of the two points on the function's graph:  \((1, f(1))\)   and  \((3, f(3))\).

• Calculate  \(f(1)\):
   \(f(1) = (1)^2 - 2(1) + 2 = 1 - 2 + 2 = 1\)
• Calculate  \(f(3)\):
   \(f(3) = (3)^2 - 2(3) + 2 = 9 - 6 + 2 = 5\)

So, the two points are  \((1, 1)\)   and  \((3, 5)\). Now, we calculate the slope of the secant line joining these two points using the formula  \(m = \frac{y_2 - y_1}{x_2 - x_1}\):

\[\text{Slope of secant line} = \frac{f(3) - f(1)}{3 - 1} = \frac{5 - 1}{2} = \frac{4}{2} = 2\]

The slope of the secant line is  \(2\).

Finding the Derivative and Tangent Slope

Next, we need to find the slope of the tangent line to  \(f(x)\)   at any point  \(x\). This is given by the derivative  \(f'(x)\).

Given  \(f(x) = x^2 - 2x + 2\), its derivative is:

\[f'(x) = \frac{d}{dx}(x^2 - 2x + 2)\]

\[f'(x) = 2x - 2\]

This expression represents the slope of the tangent line at any point  \(x\).

Solving for the Point x

According to the Mean Value Theorem and the problem statement, the tangent line at the desired point  \(x\)   must be parallel to the secant line. This means their slopes must be equal.

We set the slope of the tangent  \(f'(x)\)   equal to the slope of the secant line (which we found to be  \(2\)):

\[f'(x) = \text{Slope of secant line}\]

\[2x - 2 = 2\]

Now, we solve this equation for  \(x\):

• Add  \(2\)   to both sides:
   \(2x = 2 + 2\)
•  \(2x = 4\)
• Divide by  \(2\):
   \(x = \frac{4}{2}\)
•  \(x = 2\)

Finally, we check if this value of  \(x\)   lies within the given interval  \([1, 3]\). Since  \(1 \le 2 \le 3\), the value  \(x = 2\)   is indeed within the interval.

Summary of Steps and Result

Here's a concise summary of the process:

The point  \(x\)   at which the tangent of  \(f(x)\)   becomes parallel to the straight line joining  \(f(1)\)   and  \(f(3)\)   is  \(2\).