Question

a. Use definition ( 2 ) ( \(p .\) 135 ) to find the slope of the line tangent to the graph of \(f\) at \(P\). b. Determine an equation of the tangent line at \(P\). $$f(x)=2 x+1 ; P(0,1)$$

Short answer

In the problem, we are given a function \(f(x) = 2x + 1\) and a point \(P(0,1)\). We are asked to find the slope of the line tangent to the graph of \(f\) at point \(P\) and determine an equation for that tangent line. After following the steps provided in the solution, we find that the derivative of the function is \(f'(x) = 2\). This allows us to conclude that the slope of the tangent line at any point on the graph of \(f\) is 2, including the slope at point \(P(0,1)\). Using the point-slope equation, we determine the equation of the tangent line to be \(y = 2x + 1\).

Step-by-step solution

Find the derivative of f(x)

Using the definition of the derivative, we compute \(f'(x)\): $$f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$ Let's calculate this limit for the given function: $$f'(x) = \lim_{h\to0} \frac{(2(x+h)+1)-(2x+1)}{h}$$

Simplify the expression

Now, simplify the expression in the limit: $$f'(x) = \lim_{h\to0} \frac{2x+2h+1-2x-1}{h} =\lim_{h\to0} \frac{2h}{h}$$

Evaluate the limit

Divide out the \(h\) and evaluate the limit: $$f'(x) = \lim_{h\to0} 2 = 2$$ So the derivative of \(f(x)\) is a constant, \(f'(x) = 2\). This means that the slope of the tangent line at any point on the graph of \(f\) is 2.

Find the slope at point P

We found that the slope of the tangent line at any point on the graph is 2, so the slope at point \(P(0,1)\) must also be 2.

Determine the equation of the tangent line using point-slope form

Now, we're going to use point-slope form to find the equation of the tangent line: $$y - y_1 = m(x - x_1)$$ Where \(m\) is the slope we found in Step 4, and \((x_1, y_1)\) is the given point \(P(0,1)\). Plugging these values into the formula gives us: $$y - 1 = 2(x - 0)$$ Simplify the equation to get the final result: $$y = 2x + 1$$ So, the equation of the tangent line is \(y = 2x + 1\).

Key concepts

Derivative Calculus

In calculus, derivatives capture the concept of how one quantity changes in response to changes in another quantity. The derivative of a function at a point is essentially the slope of the line tangent to the function's graph at that point. It gives us the rate at which the function's value is changing at that specific point.

For example, in the given exercise, the function \(f(x) = 2x + 1\) is a straight line, and we can intuitively see that its steepness does not change no matter where we are on the line. Using calculus, we formalize this by finding the derivative, which tells us the exact steepness of the line at every point. In this case, the derivative is a constant \(f'(x) = 2\), showing that the function has a constant rate of change – its slope – everywhere.

Limit Definition of Derivative

The limit definition of the derivative is a foundational concept that allows us to calculate how a function changes at any given point. Derivatives describe this change mathematically as the limit of the average rate of change (the difference quotient) as the interval between the points shrinks to zero.

The formula given by the limit definition is: \[ f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \]As shown in the step-by-step solution for our function \(f(x)\), by applying this definition, we can systematically determine the derivative regardless of our intuitive understanding of the graph.

Slope of Tangent

The slope of a tangent line to the graph of a function at a particular point represents how fast the function is increasing or decreasing at that point. If the slope is positive, the function is rising, and if the slope is negative, the function is falling. For linear functions, the slope is the same at every point on the graph.

In our exercise, since \(f(x)\) is linear, the slope of the tangent at any given point \(P\) is the derivative \(f'(x)\). For \(P(0,1)\), the slope of the tangent \(m\) is the value of the derivative at \(x=0\), which is \(f'(0) = 2\). This constant slope indicates that the function maintains a uniform ascent across its entire domain.

Point-Slope Form

The point-slope form is a method for finding the equation of a line if you know the slope and a single point on the line. It’s written as \[ y - y_1 = m(x - x_1) \] where \(m\) represents the slope of the line and \( (x_1, y_1) \) is a known point on the line.

Using the calculations from our earlier steps, where we determined the slope \(m = 2\) and have the point \(P(0,1)\), we can create the equation of our tangent line simply by plugging these values into our point-slope formula. The final equation represents a line that just grazes the graph of \(f\) at point \(P\), never crossing it, which is the exact property of a tangent.