Question

At which points on the curve y = 1 + 40x3 − 3x5 does the tangent line have the largest slope?

(x,y) Smaller x-value
(x,y) Larger x-value

Answer 1

To find the points on the curve where the tangent line has the largest slope, we need to find the maximum value of the derivative of the curve.

Step-by-step explanation:

To find the points on the curve where the tangent line has the largest slope, we need to find the maximum value of the derivative of the curve. The derivative of the curve y = 1 + 40x^3 - 3x^5 is given by y' = 120x^2 - 15x^4. To find the maximum value of the derivative, we set y' equal to 0 and solve for x.

120x^2 - 15x^4 = 0

15x^2(8 - x^2) = 0

x = 0 or x = ±√8

Since we are looking for the points with smaller x-values and larger x-values, the two points are (0, 1) and (√8, -15).

Answer 2

To find where the tangent line has the largest slope on the curve y = 1 + 40x³ − 3x⁵, we calculate the function's derivative, set it equal to zero to find the critical points, and then apply the second derivative test to determine maxima. The slope is largest at x = -√8 and x = √8.

To find the points on the curve where the tangent line has the largest slope at the smaller and larger x-values, we'll need to work with the first derivative of the function.

Given
`\(y = 1 + 40x^3 - 3x^5\)`, the slope of the tangent line to a curve at any point is given by the derivative of y with respect to x,
`(dy)/(dx)`.

Let's first find the derivative of y with respect to x:

`\[y = 1 + 40x^3 - 3x^5\]`

`\[(dy)/(dx) = (d)/(dx) (1) + (d)/(dx) (40x^3) - (d)/(dx) (3x^5)\]`

`\[(dy)/(dx) = 0 + 120x^2 - 15x^4\]`

To find the points where the tangent line has the largest slope, we'll need to find the critical points of the derivative, i.e., where \(dy/dx = 0\) or where the derivative is undefined.

`\[(dy)/(dx) = 120x^2 - 15x^4\]`

Setting the derivative equal to zero to find critical points:

`\[120x^2 - 15x^4 = 0\]`

`\[x^2(120 - 15x^2) = 0\]`

This equation gives us two critical points:

1.
`\(x^2 = 0\) \rightarrow \(x = 0\) (since \(x^2 = 0\)` has a solution at
`\(x = 0\))`

2.
`\(120 - 15x^2 = 0\) \rightarrow \(15x^2 = 120\) \rightarrow \(x^2 = 8\) \rightarrow\alpha \(x = √(8)\) or \(x = -√(8)\)`

Now, to determine which critical point corresponds to the largest slope at smaller and larger x-values, we'll evaluate the second derivative or check the slopes at these points.

`\[(d^2y)/(dx^2) = (d)/(dx)(120x^2 - 15x^4)\]`

`\[(d^2y)/(dx^2) = 240x - 60x^3\]`

Now, substitute the critical points (x = 0) and
`\(x = √(8)\)` into the second derivative:

1. For (x = 0):

`\[d^2y/dx^2 = 240(0) - 60(0)^3 = 0\]`

2. For \(x = \sqrt{8}\):

`\[d^2y/dx^2 = 240(√(8)) - 60(√(8))^3\]`

`\[d^2y/dx^2 = 240√(8) - 60(8√(8))\]`

`\[d^2y/dx^2 = 240√(8) - 480√(8)\]`

`\[d^2y/dx^2 = -240√(8) < 0\]`

From the second derivative test, the point
`\(x = √(8)\)` yields a negative second derivative, indicating a local maximum. Therefore, at
`\(x = √(8)\)`, the slope of the tangent line is largest among the critical points for the larger x-value.

However, since
`\(x = √(8)\) and \(x = -√(8)\)` correspond to the same absolute value of x but with opposite signs, for the smallest x-value, the slope will also be the largest at
`\(x = -√(8)\)`.