Question

f(X)=x^3-6x^2+7x-5. find the open intervals on which f is concave up and down. then determine the inflection points.

Answer 1

First step: take the first and second derivatives.

Answer 2

so 3x^2-12x+7=f'(x) then 6x-12=f"(x)?

Answer 3

Those are correct. Now, do you remember how to use those to determine concavity and inflection points?

Answer 4

if f"(x)>0 or f"(X)<0?

Answer 5

That is correct. If f''(x)>0, then it is convex (concave up), if f''(x)<0, then it is concave (concave down). Inflection point is where it switches.

Answer 6

ohhhh okay thanks!

Answer 7

inflection point
\[x^3 -6x^2 +7x -5=0\]
first derivative
\[3x^2-12x+7=0\]
second derivative
\[6x-12=0\]
x=2
y=-7 subs. x=2 to the equation
2,-7 inflection point

Answer 8

inflection point can be found in second derivative

Answer 9

it is the turning point

Answer 10

what about for f(x)=x^2+5x-14? because then the second derivative is 2x+5 and the second is just 2, so how do I find the interval on which it is concave up and down and the inflection points with just 2?

Answer 11

If the second derivative is f''(x)=2, that means that it is a constant function that is always 2. Since it is always positive, that means f'(x) is always increasing, which means f(x) is convex (concave up) over the entire real line. No inflection points.

Answer 12

okay makes sense

Answer 13

f(x)=(5x-6)/(x+4) so the second derivative is (x^2-44x-192)/(x+4)^4 so x= 48 and -4. i found that f is concave up on the interval (-inf, -4), but i can't figure out when it is concave down and the inflection points