Don't you mean f(6) = 3?
A Taylor series, expanded about x = 6, says that
f(7) = f(6) + f'(6)*1 + (1/2!)*f"(x)*1^2 + ...
= 3 -(1/2) -(2/2)
+ (higher order terms)
= 1.5 + higher order terms
With a large positive third derivative, any of the values listed could be possible.
I am also confused
The function f is twice differentiable, and the graph of f has no points of inflection. If f(6)+3, f'(6)=-1/2, and f"(6)=-2, which of th following could be the value of f(7).
A.)2
B.)2.5
C.)2.9
D.)3
E.)4
the answer is A. but what are the steps to reach this answer?
1 answer
1 answer