Problem 43 - Calculus an Applied Approach

Question

Use the limit definition to find an equation of the tangent line to the graph of \(f\) at the given point. Then verify your results by using a graphing utility to graph the function and its tangent line at the point. $$ f(x)=\sqrt{x}+1 ;(4,3) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of function \(f(x)=\sqrt{x}+1\) at point (4,3) is \(y = 0.25x + 2\)

Step 1: Identify the given function and point

The given function is \(f(x) = \sqrt{x} + 1\) and the point on which the line is tangent to the function is \((4,3)\)

Step 2: Differentiating using limit definition

The derivative of the function \(f(x)\) would give us the slope of the tangent line. The limit definition of the derivative is \[f'(x) = \lim_{h\to0} \dfrac{f(x+h) - f(x)}{h} \] So, substituting our function into the equation and simplify to get: \[f'(x) = \lim_{h\to0} \left( \frac{\sqrt{x+h} + 1 - (\sqrt{x} + 1)}{h} \right)\] This simplifies (by cancelling the 1's) to: \[f'(x) = \lim_{h\to0} \left( \frac{\sqrt{x+h} - \sqrt{x}}{h} \right)\] Using rationalization technique we get: \[f'(x) = \lim_{h\to0} \left( \frac{h}{h (\sqrt{x+h} + \sqrt{x})} \right)\] After cancelling h and as h goes to 0, the equation simplifies further to: \[f'(x) = \frac{1}{2 \sqrt{x}} \]

Step 3: Plug in the x-coordinate of the given point

We plug \(x = 4\) (the x-coordinate of the given point) into the derivative to find the slope of the tangent line at that point: \[f'(4) = \frac{1}{2 \sqrt{4}} = 0.25 \] So, the slope of the tangent line at point (4,3) is 0.25.

Step 4: Use the point-slope formula to find the equation of the tangent line

The point-slope formula is \(y - y1 = m(x - x1)\). Substituting our point (4,3) and slope (0.25) into this formula, we get: \[y - 3 = 0.25(x - 4)\] After distributing 0.25 and adding 3, we get the equation of the tangent line: \[y = 0.25x + 2\] This is the equation of the tangent line to the graph of the function \(f(x)\) at the point (4,3).

Step 5: Verification

The verification of this equation with a graphing utility would have to be performed. The graph should show the function \(f(x) = \sqrt{x} + 1\) and its tangent line \(y = 0.25x + 2\) at the point (4,3). Since the graphical verification cannot be shown here, this step is more of a recommendation for the next possible action.

Key Concepts

Limit Definition of the Derivative

Understanding the limit definition of the derivative is crucial for grasping how calculus describes change. At its heart, the derivative represents the best linear approximation of a function at a certain point, and hence, the slope of the tangent line at that point.

The mathematical expression for the derivative using the limit is
\[f'(x) = \lim_{{h\to0}} \dfrac{{f(x+h) - f(x)}}{{h}} \]
This can appear daunting at first, but it essentially compares the initial function value at a point 'x' to its value at a slightly different point 'x+h'. As 'h' approaches zero, the ratio offers the slope of the tangent line at 'x'. Breaking down our example, we start with function
\(f(x) = \sqrt{x} + 1\), and apply this limit process to compute the slope.

Slope of a Tangent Line

The slope of a tangent line at any point on a curve is a snapshot of the curve's steepness at that exact location. It can be found using the derivative of the function at that point. For example, given the function
\(f(x) = \sqrt{x} + 1\)
we found the derivative, which gives us the formula for the slope of the tangent line across all points of the function. However, the specific value of that slope at any point requires substituting the particular 'x' value from the point of tangency into the derivative. As we did for point
\((4,3)\), plugging the 'x' value into the slope formula
\(f'(x)\) yields the precise slope where the tangent line just kisses the curve.

Rationalization Technique

In calculus, the rationalization technique is a method to eliminate radicals from the denominator of a fraction. We often employ this method when working with the limit definition of a derivative, especially for functions containing roots, as in the case of
\(f(x) = \sqrt{x} + 1\).

After applying the limit definition, we are left with an expression that contains a difference of square roots in the denominator. To simplify it, we multiply both the numerator and the denominator by the conjugate of the denominator; in essence, we create a fraction that allows us to apply the well-known difference of squares. This process clears the radical and allows us to cancel terms, providing a clearer path to the limit and, consequently, the slope of the tangent line.

Point-Slope Formula

Once we know the slope of the tangent line and a specific point of tangency, we turn to the point-slope formula to write the equation of this line. This formula is
\(y - y_1 = m(x - x_1)\),
where \(m\) represents the slope, and \((x_1, y_1)\) is the given point on the line.

The beauty of the point-slope formula is its directness. In our scenario, for the point
\((4,3)\) with a slope of 0.25, we simply plug these values into the formula to produce
\(y - 3 = 0.25(x - 4)\).
It’s a hassle-free approach to churn out the linear equation that accurately describes the tangent line at the specific point on the curve, completing our goal of finding the tangent line's equation.