f (x) = 5x

f(x) = 5x.dx = (5/4)x C

since 135x y = 0 => at P(x,y) on f(x)

y = -135x ( cp. y = mx c)

grad = f'(x) = -135

5x = -135 => x = (-27) = -3

135(-3) y =0 => y = 405

level of contact at tangent of f(x) is P(-3,405)

y= (5/4)x C =>

405 = (5/4)(-3) C

C = 405 405/4 = 3*405/4 = 1215/4

Supply(s): Math main

=> f(x) = (5/4)[x 243]

To undo a by-product, discover the integral. The integral of 5x^3 is 5/4x^4 C. Thats the type of the equation. To search out C, we should keep in mind that the slope of the tangent line is the by-product of the operate at that time. We are able to rewrite 135x y = 0 as y = -135x. So the slope is -135 sooner or later x. That implies that the by-product at that time is -135. 5x^3 = -135. Divide either side by 5 and also you get x^3=-27. So x is -3 the place the tangent line is positioned. That implies that y at that time is -135(-3) = 405. So, plug each factors in for the equation of the operate. 405 = 5/4(3)^4 C. Simplify: 405 = 101.25 C; C = 506.25. The equation is then f(x) = 5/4x^4 506.25.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022