Question

Suppose the line tangent to the graph of \(f\) at \(x=2\) is \(y=4 x+1\) and suppose the line tangent to the graph of \(g\) at \(x=2\) has slope 3 and passes through (0,-2) . Find an equation of the line tangent to the following curves at \(x=2\) a. \(y=f(x)+g(x)\) b. \(y=f(x)-2 g(x)\) c. \(y=4 f(x)\)

Short answer

a. y=f(x)+g(x) b. y=f(x)-2g(x) c. y=4f(x) #tag_title#Answer#tag_content#a. y=7x-1 b. y=-2x+5 c. y=16x-4

Step-by-step solution

Identify given information and goal

We are given the following information: - For function \(f(x)\): The tangent line at x=2 is \(y=4x+1\). - For function \(g(x)\): The tangent line at x=2 has a slope of 3 and passes through (0,-2). Our goal is to find the equations of the tangent lines for the following functions at x=2: a. \(y=f(x)+g(x)\) b. \(y=f(x)-2g(x)\) c. \(y=4f(x)\)

Find the tangent point for f(x) and g(x) at x=2

First, we need to find the points of tangency for f(x) and g(x) at x=2. For \(f(x)\), we are given the tangent line equation \(y=4x+1\). Plug in x=2: \(y=4(2)+1 = 9\) So the tangent point for \(f(x)\) at x=2 is \((2,9)\). For \(g(x)\), we know the tangent line has a slope of 3 and passes through (0,-2). Use the point-slope form of a linear equation to find the equation of the tangent line of \(g(x)\): \(y-(-2)=3(x-0) \Rightarrow y=3x-2\) Now plug in x=2: \(y=3(2)-2 = 4\) So the tangent point for \(g(x)\) at x=2 is (2, 4).

Find the tangent lines for the new functions

Now that we have the tangent points for \(f(x)\) and \(g(x)\), we can find the tangent lines for the new functions at x=2. a. \(y=f(x)+g(x)\) Slope of the tangent line at x=2: \(4+3=7\) Tangent point: \((2,9+4) = (2,13)\) Use the point-slope form: \(y-13=7(x-2)\). Simplify the equation: \(y=7x-1\) b. \(y=f(x)-2g(x)\) Slope of the tangent line at x=2: \(4-2(3)= -2\) Tangent point: \((2,9-2(4)) = (2,1)\) Use the point-slope form: \(y-1=-2(x-2)\). Simplify the equation: \(y=-2x+5\) c. \(y=4f(x)\) Slope of the tangent line at x=2: \(4(4)=16\) Tangent point: \((2,4(9)) = (2,36)\) Use the point-slope form: \(y-36=16(x-2)\). Simplify the equation: \(y=16x-4\) The equations of the tangent lines for the given functions are: a. \(y=7x-1\) b. \(y=-2x+5\) c. \(y=16x-4\)

Key concepts

Differential Calculus

Differential calculus is a subfield of calculus that focuses on the concept of change and how functions vary. One of its fundamental tools is the derivative, which measures the rate at which a function's output changes as its input changes. More simply, it gives us the slope of the tangent line to the curve of the function at a given point.

In the context of our exercise, differential calculus is used to determine the slope of the tangent lines to the functions at a specific point, here at x=2. Knowing the slope allows us to write the equation of the tangent line, which will be a linear function that just touches the original curve at that point without crossing it.

Tangency Points

Tangency points, often referred to simply as points of tangency, are the exact spots where a tangent line touches a curve. At these points, the tangent line has the same instantaneous direction as the curve, meaning they share the same slope at that precise location.

For the functions f(x) and g(x) in our exercise, we identified the tangency points at x=2 by using the given tangent line equations and solving for the corresponding y-values. These points are crucial in constructing the tangent lines for any new functions derived from f(x) and g(x), such as f(x) + g(x), f(x) - 2g(x), and 4f(x). They provide us with the anchoring coordinates that, along with the slopes, define the tangent lines.

Slope of Tangent

The slope of the tangent line to a curve at any given point is a central concept in differential calculus. This slope is actually the value of the derivative of the function at that specific point. It indicates the steepness of the line and defines its angle with respect to the horizontal axis.

For the given exercise, we use the known slopes of the tangents to functions f(x) and g(x) to find the slopes of the tangents to our new functions at x=2. For instance, knowing the slope of f(x) is 4 and g(x) is 3 at x=2, we can calculate the slope for f(x) + g(x) by adding these slopes to get 7. This addition is due to the fact that the derivative of a sum of functions is the sum of their derivatives. Similar operations are performed for subtraction and scalar multiplication to find the tangent slopes for f(x) - 2g(x) and 4f(x), respectively.