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Answer :
The tangent line has its largest slope at [tex]\( x = \sqrt{18} \)[/tex]. The point with the smallest[tex]\( x \)[/tex]-value is [tex]\( (0, 1) \),[/tex] and the largest [tex]\( x \)[/tex]-value is approximately [tex]\( \left(\sqrt{18}, 1 + 432\sqrt{2}\right) \).[/tex]
To find where the tangent line has the largest slope on the curve [tex]\( y = 1 + 60x^3 - 2x^5 \),[/tex] we need to find where the derivative of the curve is maximized.
The derivative gives the slope of the tangent line at any point on the curve.
The derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ y' = \frac{dy}{dx} = 180x^2 - 10x^4 \][/tex]
To find the maximum slope, we set the derivative equal to zero and solve for [tex]\( x \):[/tex]
[tex]\[ 180x^2 - 10x^4 = 0 \][/tex]
[tex]\[ x^2(180 - 10x^2) = 0 \][/tex]
This gives us two critical points: [tex]\( x = 0 \) and \( x = \sqrt{18} \).[/tex]
1. At[tex]\( x = 0 \)[/tex], the slope is 0, indicating a local minimum or maximum.
2. At [tex]\( x = \sqrt{18} \)[/tex], the slope is maximized.
To find the corresponding [tex]\( y \)-[/tex]values at these critical points:
1. [tex]\( x = 0 \):[/tex]
[tex]\[ y = 1 + 60(0)^3 - 2(0)^5 = 1 \][/tex]
So, the point is [tex]\( (0, 1) \).[/tex]
[tex]2. \( x = \sqrt{18} \):[/tex]
[tex]\[ y = 1 + 60(\sqrt{18})^3 - 2(\sqrt{18})^5 \][/tex]
[tex]\[ y ≈ 1 + 60 \times 18\sqrt{18} - 2 \times 18^2 \sqrt{18} \][/tex]
[tex]\[ y ≈ 1 + 60 \times 18\sqrt{2} - 2 \times 18^2 \sqrt{2} \][/tex]
[tex]\[ y ≈ 1 + 1080\sqrt{2} - 648\sqrt{2} \][/tex]
[tex]\[ y ≈ 1 + 432\sqrt{2} \][/tex]
So, the point is approximately [tex]\( \left(\sqrt{18}, 1 + 432\sqrt{2}\right) \).[/tex]
Therefore, the point with the smallest[tex]\( x \)-[/tex]value is [tex]\( (0, 1) \),[/tex]and the point with the largest [tex]\( x \)[/tex]-value is approximately [tex]\( \left(\sqrt{18}, 1 + 432\sqrt{2}\right) \).[/tex]
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