Answer:
0
Step-by-step explanation:
Put -6 where x is and do the arithmetic.
f(-6) = 1/3 (-6 +6) = 1/3 (0)
f(-6) = 0
ANSWER
B.Yes, f is continuous on [1, 7] and differentiable on (1, 7).

EXPLANATION
The given

The hypotheses are
1. The function is continuous on [1, 7].
2. The function is differentiable on (1, 7).
3. There is a c, such that:


This implies that;




Since the function is continuous on [1, 7] and differentiable on (1, 7) it satisfies the mean value theorem.
All you have to use is pemdas and keep y by itself
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