Question

How to solve this question?

Answer 1

Answer:

which there is not question

Answer 2

Given:  The curve $y=ax^{3} -4x +1$ is parallel to line $y= 50x-1$ at $x=3$

To find: (I) Gradient, m at  $x=4$,

             (II) Tangent at $x=-3$ is parallel to line $y=50x-1$

Solution:   $x=3$

on comparing the line $y= 50x-1$ with $y=mx+c$, slope m = 50

$m=$ $\frac{dy}{dx}$$=\frac{d}{dx}(ax^{3}-4x+1 )$ $=$ $3ax^{2} -4$  -----(1) , since x= 3 given

                                         $27a-4=50$⇒ $a=2$, (since m=50)

from equation (1) $m=3X2Xx^{2}$$-4$ $=6x^{2}$$-4$

(I) at $x=4$, $m=6X(4)^{2}$$-4$  $=$   $96-4=92$

(II)at $x=-3, m= 6x^{2} -4=6(-3)^{2} -4=6X9-4=54-4=50$

so slope at $x=-3$ is equal to the slope of the line $y= 50x-1$,

hence tanget at x = -3 is parallel to the line y = 50x - 1,

hence proved.