Problem 75 - Algebra and Trigonometry

Question

The slope \(m\) of the tangent line to the graph of \(f(x)=3 x^{4}-7 x^{2}+2\) at any number \(x\) is given by \(m=f^{\prime}(x)=12 x^{3}-14 x\). Find an equation of the tangent line at \(x=1\).

Step-by-Step Solution

Answer

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

Step 1 - Find the derivative

The derivative of the function, which gives the slope of the tangent line, is provided as: \[ f^{\text{ '} }(x) = 12x^{3} - 14x \]

Step 2 - Calculate the slope at x=1

Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at this point:\[ m = f^{\text{ '} }(1) = 12(1)^{3} - 14(1) = 12 - 14 = -2 \]

Step 3 - Evaluate the function at x=1

Substitute \( x = 1 \) into the original function to find the y-coordinate:\[ f(1) = 3(1)^{4} - 7(1)^{2} + 2 = 3 - 7 + 2 = -2 \]

Step 4 - Use point-slope form

The equation of the tangent line can be found using the point-slope form: \[ y - y_{1} = m(x - x_{1}) \]where \( m = -2 \), \( x_{1} = 1 \), and \( y_{1} = -2 \).

Step 5 - Write the equation of the tangent line

Substitute the slope and the point \( (1, -2) \) into the point-slope form equation: \[ y - (-2) = -2(x - 1) \]Simplify to get the final equation of the tangent line: \[ y + 2 = -2(x - 1) \] \[ y + 2 = -2x + 2 \] \[ y = -2x + 2 - 2 \] \[ y = -2x \]

Key Concepts

Derivative

In calculus, the derivative of a function captures the rate at which the function's value changes as its input changes. It is essentially the slope of the function at any given point. For our function, we have: \[ f(x)=3x^4 - 7x^2 + 2 \] The derivative of this function is given as: \[ f'(x) = 12x^3 - 14x \] This means that the slope of the graph of our function at any point is represented by the value of this derivative. The process of finding the derivative involves applying differentiation rules to each term of the function.

Tangent Line

A tangent line to a graph at a specific point is a straight line that just touches the graph at that point. The slope of this tangent line is the same as the slope of the function at that point. In this case, to find the tangent line at \( x=1 \), we first need to determine the slope of the function at \( x=1 \). Once we know the slope, we can draw the tangent line by using the equation of the line that passes through the given point on the graph with that slope.

Point-Slope Form

The point-slope form is a way to write the equation of a line when you know the slope of the line and one point on it. The formula for the point-slope form is: \[ y - y_1 = m(x - x_1) \] Here, \( m \) is the slope, and \( (x_1, y_1) \) are the coordinates of the given point. For our task, once we have the slope \( m = -2 \) and the point \( (1, -2) \), we can plug these values into the point-slope form equation to find the tangent line.

Function Evaluation

To find the y-coordinate of the point where the tangent line touches the function's graph, we need to evaluate the function at the given x-coordinate. For \( x=1 \), substitute \( x \) into the original function: \[ f(1) = 3(1)^4 - 7(1)^2 + 2 = 3 - 7 + 2 = -2 \] This tells us that the function's value at \( x=1 \) is \( -2 \), giving us the point \( (1, -2) \) on the graph.

Slope Calculation

The slope of the tangent line at a given point is the value of the derivative at that point. To find this slope for \( x=1 \), we substitute \( x=1 \) into the derivative: \[ f'(1) = 12(1)^3 - 14(1) = 12 - 14 = -2 \] This slope tells us how steep the tangent line is at \( x=1 \). With a slope of \( -2 \), it means the curve of the function is decreasing at this rate at \( x=1 \).