Help!!! ( look at picture)
2 answers:
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The equation of the tangent line at x=1 can be written in point-slope form as
... L(x) = f'(1)(x -1) +f(1)
The derivative is ...
... f'(x) = 4x^3 +4x
so the slope of the tangent line is f'(1) = 4+4 = 8.
The value of the function at x=1 is
... f(1) = 1^4 +2·1^2 = 3
So, your linearization is ...
... L(x) = 8(x -1) +3
or
... L(x) = 8x -5
Eq. of line through (1,2) and (4,5) is
x-y+1=0
x-y+1=0