The maximum slope of the curve [y = 1 2x^4 - 5x^3 + 18x^2 - 19x] occurs at the point :

Question

The maximum slope of the curve $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$ occurs at the point :

Accepted Answer

Given, $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$

$${{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19$$

$$ \Rightarrow $$ Slope M = $2{x^3} - 15{x^2} + 36x - 19$

At max of slope $${{dM} \over {dx}} = 0$$

$$ \therefore $$ $${{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0$$

$$ \Rightarrow 6({x^2} - 5x + 6) = 0$$

$$ \Rightarrow 6(x - 2)(x - 3) = 0$$

$$ \therefore $$ $x = 2,3$

Now, $${{{d^2}M} \over {d{x^2}}} = 6(2x - 5)$$

at $$x = 2,{{{d^2}M} \over {d{x^2}}} = 6(4 - 5) =  - 6 < 0$$

$$ \therefore $$ at x = 2 slope is maximum.

At x = 2,

y = 8 - 40 + 72 - 38 = 2

$$ \therefore $$ Required point = (2, 2)

Solutions · 1