Question

How do you find the tangents to the curve \[y = {x^3} + x\] at the points where the slope is 4?

Answer 1

Hint: To find the tangents to the curve \[y = {x^3} + x\] at the points, we need to find the points of the curve using power rule in which; after applying this we need to substitute the points obtained in point slope form formula i.e., \[y - {y_1} = m\left( {x - {x_1}} \right)\] ;to find the tangents of the given curve with the given slope.
Formula used:
 \[y - {y_1} = m\left( {x - {x_1}} \right)\]
 \[y\] = y-coordinate of second point.
 \[{y_1}\] = y-coordinate of point one.
 \[m\] = slope.
 \[x\] = x-coordinate of second point.
 \[{x_1}\] = x-coordinate of point one.

Complete step by step solution:
Let us write the given curve:
 \[y = {x^3} + x\] …………………. 1
When the tangent line to a function has a slope of 4, this is the same as saying that the derivative of the function is equal to 4 at that same point.
Thus, we have to find the derivative of the function and set it equal to 4.
To differentiate \[y = {x^3} + x\] , we will use the Power rule. Through the power rule, we get:
 \[\dfrac{{dy}}{{dx}} = 3{x^2} + 1\] .
 So, we then set \[\dfrac{{dy}}{{dx}} = 4\] ; since this is the slope of the tangent line at the desired points. This gives us the equation as:
 \[3{x^2} + 1 = 4\]
 \[ \Rightarrow 3{x^2} = 4 - 1\]
Evaluating the terms, we get:
 \[3{x^2} = 3\]
Now, divide both sides of the equation by 3 as:
 \[\dfrac{{3{x^2}}}{3} = \dfrac{3}{3}\]
Hence, we get:
 \[ \Rightarrow {x^2} = \dfrac{3}{3} = 1\]
 \[ \Rightarrow x = \pm 1\]
Which implies that \[x = \pm 1\] . Thus, we have two points. Determine their coordinates by plugging \[\left( { - 1,1} \right)\] into the equation 1 for y as we have:
 \[y = {x^3} + x\]
 \[y\left( { - 1} \right) = {\left( { - 1} \right)^3} + \left( { - 1} \right)\]
 \[ \Rightarrow y\left( { - 1} \right) = - 2\]
Hence, we get the points as \[\left( { - 1, - 2} \right)\] .
Now, for \[y\left( 1 \right)\] we have:
 \[y\left( 1 \right) = {\left( 1 \right)^3} + 1\]
 \[ \Rightarrow y\left( 1 \right) = 2\]
Hence, we get the points as \[\left( {1,2} \right)\] .
To determine the equations of the tangent lines, plug the points and slope of 4 into equations in point-slope form, which takes the point \[\left( {{x_1},{y_1}} \right)\] and slope m and relates them as the equation:
 \[y - {y_1} = m\left( {x - {x_1}} \right)\] …………………… 2
Thus, we have two equations i.e., for \[\left( { - 1, - 2} \right)\] we have \[{x_1},{y_1} = - 1, - 2\] and for \[\left( {1,2} \right)\] we have \[{x_1},{y_1} = 1,2\] and Slope: \[m = 4\] .
Now, let us solve the equation with \[{x_1},{y_1} = - 1, - 2\] using equation 2 as:
 \[y - {y_1} = m\left( {x - {x_1}} \right)\]
 \[y - \left( { - 2} \right) = 4\left( {x - \left( { - 1} \right)} \right)\]
Simplifying the terms, we get:
 \[y + 2 = 4\left( {x + 1} \right)\]
 \[ \Rightarrow y = 4x + 2\]
Now, let us solve the equation with \[{x_1},{y_1} = 1,2\] using equation 2 as:
 \[y - {y_1} = m\left( {x - {x_1}} \right)\]
 \[y - 2 = 4\left( {x - 1} \right)\]
Simplifying the terms, we get:
 \[y = 4x - 4 + 2\]
 \[ \Rightarrow y = 4x - 2\]
So, the correct answer is “y = 4x - 2 AND y = 4x + 2”.

Note: The key point to solve this question is that we must know how to apply power rules and differentiate the terms of the given curve to obtain the coordinates, and find the point slope form. And you must know, the power rule of differentiation is defined as the expression which is in the form of \[{x^n}\] and the power is multiplied with the expression such that the power is reduced by 1.