Question

a. For the following functions, find \(f^{\prime}\) using the definition. b. Determine an equation of the line tangent to the graph of \(f\) at \((a, f(a))\) for the given value of \(a\) $$f(x)=\frac{1}{x} ; a=-5$$

Short answer

Answer: The equation of the tangent line is \(y + \frac{1}{5} = -\frac{1}{25}(x + 5)\).

Step-by-step solution

Calculate the derivative of the function using the definition of the derivative

To find the derivative, we use the definition of the derivative: $$f^{\prime}(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ We are given the function \(f(x) = \frac{1}{x}\), so we need to calculate \(f(x + h)\). To do this, we replace \(x\) with \(x + h\) in the function: $$f(x+h)=\frac{1}{x+h}$$ Now, we need to plug the function values back into the definition of the derivative.

Plug the function values into the definition of the derivative

Now that we have \(f(x) = \frac{1}{x}\) and \(f(x + h) = \frac{1}{x + h}\), we can rewrite the expression for the derivative as: $$f^{\prime}(x)=\lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$

Simplify the expression for the derivative

We can simplify the expression for the derivative using algebraic manipulations. First, find a common denominator for the fractions in the numerator: $$f^{\prime}(x)=\lim_{h \to 0} \frac{\frac{x-x-h}{x(x+h)}}{h}$$ Now, simplify the numerator: $$f^{\prime}(x)=\lim_{h \to 0} \frac{-h}{x(x+h)h}$$ Cancel out the common factor of \(h\) in the numerator and the denominator: $$f^{\prime}(x)=\lim_{h \to 0} \frac{-1}{x(x+h)}$$ Now, we can take the limit as \(h\) approaches \(0\).

Take the limit as h approaches 0

Substitute \(h = 0\) into the expression for the derivative: $$f^{\prime}(x)=\frac{-1}{x^2}$$

Find the tangent line using the point-slope form of a linear equation

Given \(a = -5\), we can find the derivative at that point using the formula we derived above: $$f^{\prime}(a)=f^{\prime}(-5)=\frac{-1}{(-5)^2}=\frac{-1}{25}$$ Now let's find the coordinates of the point \((a, f(a))\): $$f(a)=f(-5)=\frac{1}{-5}=-\frac{1}{5}$$ So, the point is \((-5, -\frac{1}{5})\). Now use the point-slope form of a linear equation to find the equation of the tangent line: $$y - y_1 = m(x - x_1)$$ Where \((x_1, y_1) = (-5, -\frac{1}{5})\) and \(m = f^{\prime}(-5) = -\frac{1}{25}\). Plug in the values: $$y - (-\frac{1}{5}) = -\frac{1}{25}(x - (-5))$$

Simplify the equation of the tangent line

Simplify the equation to obtain the final equation of the tangent line: $$y + \frac{1}{5} = -\frac{1}{25}(x + 5)$$ The equation of the tangent line to the graph of \(f(x) = \frac{1}{x}\) at \(a = -5\) is \(y + \frac{1}{5} = -\frac{1}{25}(x + 5)\).

Key concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. At the point of tangency, the tangent line has the same slope as the curve. This slope represents how steep the curve is at that particular point. Visualize a curve, like the graph of a function. The tangent line gives us a way to approximate the curve near a specific point. It can help predict the behavior of the function in that region. The tangent line is crucial when analyzing graphs because it gives a precise representation of slope or rate of change at a given point. This is especially helpful in calculus where understanding how functions change is key. In the context of our exercise, with the function given as \(f(x) = \frac{1}{x}\), the tangent line is the line that just grazes the curve at the given point \((a, f(a))\), in this case at \(a = -5\). The equation of this tangent line was found using derivatives and the point-slope form.

Limit Definition of Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. The derivative of a function at a point can be understood as the slope of the tangent line at that point. But how do we actually find this slope? That's where the limit definition of the derivative comes in.The limit definition of the derivative is given by \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]This formula computes the derivative by seeing how much \(f\) changes from \(x\) to \(x+h\), and then dividing by \(h\), as \(h\) gets infinitely small. Essentially, we are finding the slope of the secant line that becomes the tangent line as \(h\) approaches zero.For example, given \(f(x) = \frac{1}{x}\), the derivative was calculated using this definition by substituting \(f(x + h) = \frac{1}{x+h}\). Through a series of algebraic steps, we simplified the expression and took the limit as \(h\) approaches 0, yielding \(f'(x) = -\frac{1}{x^2}\). This derivative tells us the slope of the tangent line at any point \(x\) on the curve.

Point-Slope Form

The point-slope form is a method used to find the equation of a straight line when you know the slope and one point the line passes through. It's expressed as:\[ y - y_1 = m(x - x_1) \]Where:

• \((x_1, y_1)\) is the known point on the line,
• \(m\) is the slope of the line.

This form is particularly helpful in calculus when determining the equation of the tangent line. Given the slope from the derivative and the point from the function value, writing down the tangent line becomes straightforward.In our exercise, the derivative gave us the slope \(-\frac{1}{25}\) at the point \((-5, -\frac{1}{5})\). By substituting these values into the point-slope form, we derived the tangent line equation:\[ y + \frac{1}{5} = -\frac{1}{25}(x + 5) \]This equation represents the line that 'hugs' the curve of \(f(x) = \frac{1}{x}\) at \(x = -5\), providing a linear approximation of the function at that point.