College

We appreciate your visit to At which points on the curve tex y 1 60x 3 2x 5 tex does the tangent line have the largest slope tex begin array. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

At which points on the curve [tex]y = 1 + 60x^3 - 2x^5[/tex] does the tangent line have the largest slope?

[tex]
\begin{array}{ll}
(x, y) = (\square) & \text{(smaller } x \text{-value)} \\
(x, y) = (\square) & \text{(larger } x \text{-value)}
\end{array}
[/tex]

Answer :

We are given the curve

[tex]$$
y = 1 + 60x^3 - 2x^5.
$$[/tex]

To find the points on the curve where the tangent line has the largest slope, we first determine the derivative, which represents the slope of the tangent line at any point.

Step 1. Find the derivative

Differentiate the function with respect to [tex]$x$[/tex]:

[tex]$$
y' = \frac{d}{dx}\left(1 + 60x^3 - 2x^5\right) = 180x^2 - 10x^4.
$$[/tex]

This can be factored as

[tex]$$
y' = 10x^2(18 - x^2).
$$[/tex]

Step 2. Find the critical points for maximum slope

To maximize the slope given by [tex]$y'$[/tex], we need to find the extrema of [tex]$y'$[/tex] itself. That is, we differentiate [tex]$y'$[/tex] with respect to [tex]$x$[/tex]:

Let

[tex]$$
f(x) = 10x^2(18 - x^2).
$$[/tex]

Differentiate using the product rule:

[tex]\[
\begin{aligned}
f'(x) &= 10 \left[\frac{d}{dx}(x^2) \cdot (18 - x^2) + x^2 \cdot \frac{d}{dx}(18 - x^2)\right] \\
&= 10 \left[2x(18 - x^2) + x^2(-2x)\right] \\
&= 10 \left[36x - 2x^3 - 2x^3\right] \\
&= 10 \left[36x - 4x^3\right] \\
&= 40x(9 - x^2).
\end{aligned}
\][/tex]

Setting [tex]$f'(x)=0$[/tex] yields:

[tex]$$
40x(9 - x^2) = 0.
$$[/tex]

This gives [tex]$x = 0$[/tex] or

[tex]$$
9 - x^2 = 0 \quad \Longrightarrow \quad x^2 = 9 \quad \Longrightarrow \quad x = \pm 3.
$$[/tex]

Thus, the candidate values for [tex]$x$[/tex] are [tex]$-3$[/tex], [tex]$0$[/tex], and [tex]$3$[/tex].

Step 3. Evaluate the slope [tex]$y'$[/tex] at the critical points

Calculate [tex]$y'$[/tex] at each candidate:

- For [tex]$x = 0$[/tex]:

[tex]$$
y'(0) = 10 \cdot 0^2 \cdot (18 - 0^2) = 0.
$$[/tex]

- For [tex]$x = 3$[/tex]:

[tex]$$
y'(3) = 10 \cdot 3^2 \cdot \left(18 - 3^2\right) = 10 \cdot 9 \cdot (18 - 9) = 10 \cdot 9 \cdot 9 = 810.
$$[/tex]

- For [tex]$x = -3$[/tex]:

[tex]$$
y'(-3) = 10 \cdot (-3)^2 \cdot \left(18 - (-3)^2\right) = 10 \cdot 9 \cdot (18 - 9) = 10 \cdot 9 \cdot 9 = 810.
$$[/tex]

Both [tex]$x = -3$[/tex] and [tex]$x = 3$[/tex] give a slope of [tex]$810$[/tex], which is the largest possible slope. The point with [tex]$x = 0$[/tex] gives a slope of [tex]$0$[/tex], so it is not a maximum.

Step 4. Find the corresponding [tex]$y$[/tex]-values on the curve

Now substitute [tex]$x = -3$[/tex] and [tex]$x = 3$[/tex] into the original equation to find the corresponding points.

- For [tex]$x = -3$[/tex]:

[tex]$$
\begin{aligned}
y(-3) &= 1 + 60(-3)^3 - 2(-3)^5 \\
&= 1 + 60(-27) - 2(-243) \\
&= 1 - 1620 + 486 \\
&= -1133.
\end{aligned}
$$[/tex]

- For [tex]$x = 3$[/tex]:

[tex]$$
\begin{aligned}
y(3) &= 1 + 60(3)^3 - 2(3)^5 \\
&= 1 + 60(27) - 2(243) \\
&= 1 + 1620 - 486 \\
&= 1135.
\end{aligned}
$$[/tex]

Final Answer

The tangent line has the largest slope at the points

[tex]$$
(x, y) = (-3, -1133) \quad \text{(smaller $x$-value)}
$$[/tex]

and

[tex]$$
(x, y) = (3, 1135) \quad \text{(larger $x$-value)}.
$$[/tex]

Thanks for taking the time to read At which points on the curve tex y 1 60x 3 2x 5 tex does the tangent line have the largest slope tex begin array. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada