Question

true or false... if f'(0)=0 and f"(x)>0, then f is decreasing on the interval (-inf,0)

Answer 1

Recall what the signum of a first derivative tells you about the function at \(x\):
\[f'(x) < 0 \Leftrightarrow f\text{ is decreasing with respect to }x.\]\[f'(x) = 0 \Leftrightarrow f\text{ is constant with respect to }x.\]\[f'(x) > 0 \Leftrightarrow f\text{ is increasing with respect to }x.\]
Additionally recall what the signum of a second derivative tells you about the function at \(x\):
\[f''(x) < 0 \Leftrightarrow f\text{ is concave down.}\]\[f''(x) = 0 \Leftrightarrow f\text{ is neither concave up or concave down.}\]\[f''(x) > 0 \Leftrightarrow f\text{ is concave up.}\]
Try to use this information to formulate your answer. Holler if you still heed help.

Answer 2

it is constant and some how concave down... that doesn't seem like that would work... my is x<0

Answer 3

so it is not decreasing
which would make the statement false

Answer 4

Well, it's not constant everywhere...only at \(x=0\).

Answer 5

I wouldn't call it constant at 0

Answer 6

so since the x<0 it is true...

Answer 7

Of all missy ft's true/false questions, this is the most interesting, that's for sure.

Answer 8

and thanks to you jamesj i'm finally learning my calculus!!!

Answer 9

Fine, Zarkon... "Linearly approximated by a constant function at \(x=0\)".