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Home » At what point, the slope of the curve y = – x3 + 3×2 + 9x – 27 is maximum? Also find the maximum slope.
At what point, the slope of the curve y = – x3 + 3×2 + 9x – 27 is maximum? Also find the maximum slope.
Application of Derivatives | Class 12 | Exercise 1 | Maths | NCERT Exemplar
At what point, the slope of the curve y = – x3 + 3×2 + 9x – 27 is maximum? Also find the maximum slope.

Given, CURVE

    \[y\text{ }=\text{ }\text{ }x3\text{ }+\text{ }3x2\text{ }+\text{ }9x\text{ }\text{ }27\]

Separating the two sides w.r.t. x, we get

    \[dy/dx\text{ }=\text{ }-\text{ }3x2\text{ }+\text{ }6x\text{ }+\text{ }9\]

Let SLOPE of the CURVE

    \[dy/dx\text{ }=\text{ }z\]

Thus,

    \[z\text{ }=\text{ }-\text{ }3x2\text{ }+\text{ }6x\text{ }+\text{ }9\]

Separating the two sides w.r.t. x, we get

    \[dz/dx\text{ }=\text{ }-\text{ }6x\text{ }+\text{ }6\]

For LOCAL maxima and nearby minima,

    \[dz/dx\text{ }=\text{ }0\]

    \[-\text{ }6x\text{ }+\text{ }6\text{ }=\text{ }0\Rightarrow x\text{ }=\text{ }1\]

    \[d2z/dx2\text{ }=\text{ }-\text{ }6\text{ }<\text{ }0\text{ }Maxima\]

Putting

    \[x\text{ }=\text{ }1\]

in condition of the CURVE

    \[y\text{ }=\text{ }\left( -\text{ }1 \right)3\text{ }+\text{ }3\left( 1 \right)2\text{ }+\text{ }9\left( 1 \right)\text{ }\text{ }27\]

    \[=\text{ }-\text{ }1\text{ }+\text{ }3\text{ }+\text{ }9\text{ }\text{ }27\text{ }=\text{ }-\text{ }16\]

Most extreme SLOPE

    \[=\text{ }-\text{ }3\left( 1 \right)2\text{ }+\text{ }6\left( 1 \right)\text{ }+\text{ }9\text{ }=\text{ }12\]

Accordingly, (1, – 16) is where the SLOPE of the given CURVE is greatest and most extreme incline =

    \[12.\]

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