Question

At which points on the curve \(y = 1 + 40{x^3} - 3{x^5}\) does the tangent line have the largest slope?

Short answer

The tangent line on the curve\(y = 1 + 40{x^3} - 3{x^5}\) has the largest slope at\(\left( {4, - 511} \right)\).

Step-by-step solution

Given data

The equation of the curve is\(y = 1 + 40{x^3} - 3{x^5}\)

Maximization condition

The conditions for maxima of a function are:

\(\begin{array}{c}{\frac{{df\left( x \right)}}{{dx}}_{x = {x_{max}}}} = 0\;\;\;\;\;.....\left( 1 \right)\\{\frac{{{d^2}f\left( x \right)}}{{d{x^2}}}_{x = {x_{max}}}} < 0\;\;\;\;\;.....\left( 2 \right)\end{array}\)

Maximization of slope

The slope of the tangent to the curve is

\(\frac{{dy}}{{dx}} = 120{x^2} - 15{x^4}\)

From equation (1), the slop is maximum for

\(\begin{array}{c}{\frac{{{d^2}y}}{{d{x^2}}}_{x = {x_{\max }}}} = 0\\{\left( {240x - 60{x^3}} \right)_{x = {x_{\max }}}} = 0\\240{x_{\max }} = 60x_{\max }^3\\{x_{\max }} = 4\end{array}\)

Check equation (2)

\(\begin{array}{c}{\frac{{{d^3}y}}{{d{x^3}}}_{x = {x_{\max }}}} = 240 - 180x_{\max }^2\\ = 240 - 180 \times {4^2}\\ = - 2640\\ < 0\end{array}\)

The maxima condition satisfies. The corresponding \(y\) value is:

\(\begin{array}{c}{y_{\max }} = 1 + 40x_{\max }^3 - 3x_{\max }^5\\ = 1 + 40 \times {4^3} - 3 \times {4^5}\\ = - 511\end{array}\)

Thus, the slope maximum is at the point\(\left( {4, - 511} \right)\).