Question

Finding an Equation of a Tangent Line In Exercises \(33-38,\) find an equation of the line that is tangent to the graph of \(f\) and parallel to the given line. $$\begin{array}{ll}{\text { Function }} & {\text { Line }} \\ {f(x)=x^{3}+2} & {3 x-y-4=0}\end{array}$$

Short answer

The equations of the tangent lines are \(y = 3x\) and \(y = 3x + 4\).

Step-by-step solution

Find the derivative of the given function

The derivative of \(f(x) = x^3 + 2\) is \(f'(x) = 3x^2\). This derivative gives us the slope of the tangent line at any point on the function.

Find the slope of the given line

The given line is in the form \(Ax + By + C = 0\), so we can identify that the slope of the line is \(-A/B\). In line \(3x - y - 4 = 0\), A = 3 and B = -1, so the slope of this line also is 3.

Equating slopes

Since we know that the slope of the tangent to the function at a certain point is equal to the slope of the given line because they are parallel, we can equate \(3x^2\) and 3 to find the x-coordinate where the tangent line touches the function. Solving \(3x^2 = 3\) we find \(x = \pm 1\).

Find the y-coordinate where the tangent line touches the function

We substitute \(x = 1\) and \(x = - 1\) into the original function \(f(x) = x^3 + 2\) to find their corresponding y values. So, \(f(1) = 1^3 + 2 = 3\) and \(f(-1) = (-1)^3 + 2 = 1\). Therefore, the points where the tangent line touches the function are (1, 3) and (-1, 1).

Find the equations of the tangent lines

Noting that we have the points and the slope m –which is the same for both– we can use the point-slope form to build the equations. For point (1, 3) we get \(y - 3 = 3(x - 1)\) simplifying to \(y = 3x\). For point (-1, 1) we get \(y - 1 = 3(x + 1)\) simplifying to \(y = 3x + 4\).

Key concepts

Derivative of a Function

In calculus, the derivative of a function is a fundamental concept. It represents the rate at which a function's value changes at a particular point. To put simply, if you're moving along a curve, the derivative at any point gives you the slope of the tangent line at that point.

For a function described by an equation like \( f(x) = x^3 + 2 \), we find its derivative by applying rules of differentiation. The power rule is one of these rules, which says that the derivative of \( x^n \) is \( nx^{n-1} \). So, when we apply the power rule to this function, we get the derivative \( f'(x) = 3x^2 \). This tells us how steeply the curve is inclining or declining at any given value of \( x \).

Slope of a Line

The slope of a line is a measure of its steepness, usually denoted by \( m \). It's calculated as the ratio of the vertical change to the horizontal change between two points on the line. The formula to find the slope between points \((x_1, y_1)\) and \((x_2, y_2)\) is \( m = (y_2 - y_1) / (x_2 - x_1) \).

For a line in the equation form \( Ax + By + C = 0 \), the slope can also be determined by rearranging it into slope-intercept form, which is \( y = mx + b \), where \( m \) represents the slope. From our given line \( 3x - y - 4 = 0 \), by rearranging, we can see that the slope is -\( A/B \), which is 3, as indicated by the coefficients of \( x \) and \( y \). Knowing the slope of a line is crucial when investigating parallels, as parallel lines share the same slope.

Point-Slope Form

The point-slope form is used to write the equation of a line when we know its slope and any point on the line. The generic formula is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \((x_1, y_1)\) is the point.

Using this form is particularly useful when we want to express the equation of a tangent line. Once we've determined the slope of the tangent and the coordinates of the point where it touches the graph, we substitute these values into the point-slope formula. For example, if our tangent line touches the graph at point \((1, 3)\) with a slope of 3, the equation becomes \( y - 3 = 3(x - 1) \), which simplifies to the linear function of the tangent line.

Tangent Line Parallel to a Line

A tangent line to a curve at a certain point is parallel to another line if they both have the same slope. To find the equation of a such a tangent line, first we determine the slope of the given line, which must be equal to the slope of the tangent line since they are parallel. Next, we calculate the derivative of the function to find the slope of the tangent line at any point on the function.

After that, we equate the derivative to the slope of the given line to find the specific points on the function where these tangents exist. Then, we can use the point-slope form to write the equations of the tangent lines, after identifying the corresponding points on the curve. This connection between derivatives, slopes, and the point-slope form provides a straight path to finding the equations of tangent lines that are parallel to a given line.