Question

For the curve $y=4x^3-2x^5$, fnd all the points on the curve at which the tangent passes through the original.

Answer 1

Consider the given function:

$y=4x^3-2x^5$

$\frac{dy}{dx}=\frac{d}{dx}(4x^3-2x^5)$

$=3\ast 4x^2-5\ast 2x^4$

$po\mathrm{int}(x,y)$

$y_1=m{x}_{1}eq...1$

$y=4x^3-2x^{5}eq...2$

$m=12x^2-10x^4$

putting the value eq….(1)

$y_1=(4x^2-2x^4)x$

$y_1=(4x^3-2x^5)$ eq...3

Solving for the from (2) and (3 )

$\therefore 4{x_1}^3+10{x_1}^4$

$\therefore 4{x_1}^3-2{x_1}^5$

$[x_1=\pm 1]$

$for{x}_1=\pm 1putineq..y=4x^3-2x^5$

$x_1=+1$

$y=4{\left(1\right)}^3+2{\left(1\right)}^5=6$

$x_1=-1$

$y=4{(-1)}^3+2{(-1)}^5=-6$

$po\mathrm{int}are(1,6)(-1,-6)$

Hence this is the answer.