How do you find the equation of the line tangent to the graph of f(x) = x^4 + 2x^2?

Question

1790 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-03T22:46:15

Use differentiation and point slope equation to find equation of tangent at #(x_1, f(x_1))# as:

#y = 4x_1(x_1^2+1)x-x_1^2(3x_1^2+2)#

If #f(x)# is continuous and differentiable at #x_1#, then the equation of the line tangent to #f(x)# at #(x_1, f(x_1))# in point slope form is:

#y - f(x_1) = f'(x_1)(x - x_1)#

Add #f(x_1)# to both sides to get:

#y = f'(x_1)x + (f(x_1) - x_1f'(x))#

In our example, #f(x) = x^4+2x^2# so

#f'(x) = 4x^3+4x = 4x(x^2+1)#

and the equation of the tangent in slope intercept form is:

#y = 4x_1(x_1^2+1)x + ((x_1^4+2x_1^2) - x_1(4x_1^3+4x_1))#

#=4x_1(x_1^2+1)x-x_1^2(3x_1^2+2)#