Problem 49 - Applied Calculus

Question

Find the equation of the line tangent to the graph of \(f\) at \((1,1)\), where \(f\) is given by \(f(x)=2 x^{3}-2 x^{2}+1\).

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 2x - 1 \).

Step 1: Differentiate the Function

To find the equation of a tangent line to the function at a given point, we first need the derivative of the function, which provides the slope of the tangent line. The function given is \( f(x) = 2x^3 - 2x^2 + 1 \). Differentiate \( f(x) \) with respect to \( x \) to find \( f'(x) \):\[ f'(x) = \frac{d}{dx}(2x^3 - 2x^2 + 1) = 6x^2 - 4x \].

Step 2: Substitute x-coordinate into the Derivative

Substitute \( x = 1 \) into \( f'(x) \) to find the slope \( m \) of the tangent line at the point \( (1, 1) \):\[ f'(1) = 6(1)^2 - 4(1) = 6 - 4 = 2 \].Thus, the slope of the tangent line at \( (1, 1) \) is 2.

Step 3: Use Point-Slope Form to Find Tangent Line Equation

The equation of a line can be expressed in the point-slope form: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the given point on the line. Here, \( m = 2 \) and the point is \( (1, 1) \). Substitute these values into the point-slope form:\[ y - 1 = 2(x - 1) \].

Step 4: Simplify the Tangent Line Equation

Simplify the equation from the previous step to find the explicit equation of the tangent line:\[ y - 1 = 2x - 2 \].Add 1 to both sides to solve for \( y \):\[ y = 2x - 1 \].

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, crucial for understanding how a function behaves as its inputs change. It's essentially about finding how a function's output changes when we make small changes to the input. This rate of change is often represented as the derivative of the function.
In simple terms, if you imagine you're driving on a road laid out by a function graph, the derivative tells you how steep the road is at any given point. Here, the slope of the function at a specific point gives us the direction and steepness of that road (or function).
When working with polynomial functions like \(f(x) = 2x^3 - 2x^2 + 1\), differentiation boils down to applying the power rule. For each term \(ax^n\) of the polynomial, the derivative is found by multiplying \(n\) with \(a\) and reducing the exponent \(n\) by one. So, the derivative of \(2x^3\) is \(6x^2\) and that of \(-2x^2\) is \(-4x\). The constant \(1\) becomes \(0\) as the derivative of a constant is zero. Thus, the full derivative is \(f'(x) = 6x^2 - 4x\).

Point-Slope Form

The point-slope form is a linear equation format that allows you to write the equation of a line if you know its slope and one point on it. This is an incredibly handy tool, especially when dealing with tangent lines to curves. They’re lines that touch a curve at one exact point without crossing it, representing the instantaneous change in the function.
The equation of a line in point-slope form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a known point on the line and \(m\) is its slope. For the tangent line to the curve at the given point \((1, 1)\) with a calculated slope of \(2\), we set \(x_1 = 1\), \(y_1 = 1\), and \(m = 2\), substituting these values into the formula to get \(y - 1 = 2(x - 1)\).
This concise form sets us up perfectly to find the full equation of the tangent by solving for \(y\), rendering more insights into the behavior of the function at that point.

Derivative Calculation

Calculating the derivative is a primary step when finding a tangent line to a curve at a specific point. The derivative signifies the function's instantaneous rate of change at any given point on its graph.
To calculate the derivative for the given function \(f(x) = 2x^3 - 2x^2 + 1\), we apply the rules of differentiation, which simplify the process through familiar patterns.

• Constant Rule: The derivative of a constant is zero.
• Power Rule: Multiply the exponent by the coefficient and reduce the exponent by one.

Applying these rules, we compute the derivative as follows:

• The derivative of \(2x^3\) is \(6x^2\).
• The derivative of \(-2x^2\) is \(-4x\).
• The constant term \(+1\) vanishes as its derivative is zero.

Putting it all together, the function's derivative is \(f'(x) = 6x^2 - 4x\).
This derivative, evaluated at \(x = 1\), tells us the slope of the tangent line at the point \((1, 1)\). By substituting \(x = 1\) into the derivative, the calculation \(f'(1) = 6(1)^2 - 4(1) = 2\) confirms that the tangent line's slope is indeed \(2\).