Question

How do you find the equation of the tangent line to the curve `f(x) = 2x^2 + cos(x)` at the point where x = 0?

Answer 1

y = 1

Explanation:

To find the equation of the tangent , we require it's gradient and a point on it.
Now , f'(0) will give the value of the gradient and evaluating f(0) will give a point on the line.

`rArr f'(x) = 4x - sinx`

and f'(0) = 4(0) - sin0 = 0 hence gradient of line = 0

A line with a gradient = 0 , is a line parallel to the x-axis

Now f(0) = `2(0)^2 + cos0 = 0 + 1 = 1 rArr (0 , 1)`

Thus the tangent line has m = 0 and passes through ( 0 , 1)

The tangent at x = 0 is parallel to the x-axis and passes through all points on the plane with a y-coordinate = 1.

`rArr y = 1 " is the equation of the tangent "`
graph{(y-2x^2-cosx)(y-1)=0 [-10, 10, -5, 5]}