How do you find the linearization of #f(x) = x^4 + 5x^2# at the point a=1?

Question

Scarlett Allison · Accepted Answer

The linearization is the tangent line. (Or maybe it is more helpful to say: it is a way of thinking about and using the tangent line.)

#f(x) = x^4+5x^2# At #x=1#, we have #y = f(1) = 6# #f'(x) = 4x^3+10x# so at #x=1#, the slope of the tangent line is #m=f'(1) = 14# Equation of tangent line in point-slope form: #y-6 = 14(x-1)# Linearization at #a=1# (in a form I am used to): #L(x) = 6+14(x-1)# Free example of using the linearization #f(1.1) ~~ L(1.1) = 6+14(1.1-1) = 6+14(0.1) = 6+1.4 = 7.4# (The point on the tangent line with #x#-coordinate #1.1# has #y#-coordinate #7.4#.) The exact value is #f(1.1) = 1.7541# If you're very patient and steady of hand, you can find these two values using the graph: graph{(y-x^4-5x^2)(y-14x+8)=0 [0.1604, 1.846, 6.859, 7.7024]}

Evelyn Agard · Answer

undefined To find the linearization of $ f(x) = x^4 + 5x^2 $ at the point $ a = 1 $, follow these steps:

1. Calculate the derivative of $ f(x) $, denoted as $ f'(x) $.
2. Evaluate $ f'(x) $ at $ x = 1 $ to find the slope of the tangent line, denoted as $ m $.
3. Evaluate $ f(1) $ to find the value of $ f(x) $ at $ x = 1 $, denoted as $ f(1) $.
4. Use the point-slope form of a linear equation $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point of tangency, to write the equation of the tangent line.
5. Simplify the equation to obtain the linearization.

Let's go through the steps:

1. $ f(x) = x^4 + 5x^2 $
   $ f'(x) = 4x^3 + 10x $

2. Evaluate $ f'(x) $ at $ x = 1 $:
   $ f'(1) = 4(1)^3 + 10(1) = 4 + 10 = 14 $

3. Evaluate $ f(1) $:
   $ f(1) = (1)^4 + 5(1)^2 = 1 + 5 = 6 $

4. Use the point-slope form:
   $ y - 6 = 14(x - 1) $

5. Simplify the equation:
   $ y - 6 = 14x - 14 $
   $ y = 14x - 8 $

The linearization of $ f(x) = x^4 + 5x^2 $ at $ a = 1 $ is $ y = 14x - 8 $.