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
Answer:
see below
Step-by-step explanation:
Add 4 to find the solution:
z ≤ 7
The value 7 is included in the solution set, so there will be a solid dot at that point. Numbers less than 7 are also included in the solution set, so the number line will be shaded to the left of that solid dot.
