Question

Suppose that f is twice differentiable on an open interval I. If f' increases on I, then the graph of f is concave up.

true or false?

explain?

Answer 1

If f(x) increases on some interval, then f'(x) will be positive on that same interval

If f ' (x) increases on some interval, then f '' (x) will be positive on that same interval

So if f ' (x) increases on some interval, then f(x) will be concave up on that same interval.

Answer 2

So, if it's decreasing on the interval then its concave down

Answer 3

you got it

Answer 4

What about..

For c in I, if f '(c)=0 and f ''(c)<0 then f has a local minimum at x=c

and

If the point (c,f(c)) is a point of inflection, then f ''(c)=0

Answer 5

f '(c)=0 implies that there is an extrema at the point (c, f(c))

we don't know if this extrema is a local min, max, or a saddle point.

However, since f '' (c) < 0 in this same interval, this means f(x) is concave down

so there's a local max (NOT min) at (c, f(c)) when f ' (c) = 0 and f '' (c) < 0

Answer 6

so the statement

For c in I, if f '(c)=0 and f ''(c)<0 then f has a local minimum at x=c

is false

Answer 7

If the point (c,f(c)) is a point of inflection, then f ''(c)=0

is true because this is how you find the point of inflection: you derive f twice to get f '', then you find the roots of f ''

Answer 8

thank you!

Answer 9

np