Question

find the equation of the tangent and normal line to the curve y = x^3+e^x at x=0​

Answer 1

Solution:

here, the given curve is

y = x^3+e^x

at x = 0 the value of y will be

y = 0^3 + e^0

y = 0+1= 1

now, d y/dx = d/dx( x^3+e^x )

d y/dx = 3x^2+e^x

at x=0

d y/dx = 3*0^2+e^0

d y/dx =1

Equation of tangent at point (0,1) will be

(y -1) = -1 (x-0)

x +y -1 =0

Equation of normal at point (0,1) will be

(y -1) = 1(x -0)

x -y +1 =0