The formula of linearization is
L = f(a)+f'(a)(x-a)
We have [tex] f(x) = x^4+4x^2 [/tex]
And it is given in the question that a=1.
So f(a) = f(1) = 1+4=5
Now we need to find f'(a) and for that , first we need to find f'(x).
[tex] f'(x) = 4x^{4-1}+4(2)x^{2-1} [/tex]
[tex] f'(x)=4x^3+8x [/tex]
At a=1,
[tex] f'(1)= 4+8=12 [/tex]
Substituting the values of f(1) and f'(1) in the linearization formula
L=5+12(x-1)
L= 5+12x-12
L= 12x -7
And that's the required linearization .
