1. ## finding concavity

Trying to find the points of inflection of f(x)= (1/4)x^4+(7/3)x^3-(5/2)x^2+2x.

I got the intervals (1/3, 155/324) and (-5, -2495/12). This intervals seem a little extreme, did I compute this wrong.

Trying to find the points of inflection of f(x)= (1/4)x^4+(7/3)x^3-(5/2)x^2+2x.

I got the intervals (1/3, 155/324) and (-5, -2495/12). This intervals seem a little extreme, did I compute this wrong.

Perhaps: $\displaystyle f'(x)=x^3+7x^2-5x+2\Longrightarrow\,f''(x)=3x^2+14x-5$
Thus the second derivative vanishes at $\displaystyle x=-5\,,\,\,\frac{1}{3}$

As $\displaystyle f''(x)$ is an upwards parabola, it is positive (and thus f(x) is concave upwards) at...

Tonio

3. Originally Posted by tonio
Perhaps: $\displaystyle f'(x)=x^3+7x^2-5x+2\Longrightarrow\,f''(x)=3x^2+14x-5$
Thus the second derivative vanishes at $\displaystyle x=-5\,,\,\,\frac{1}{3}$

As $\displaystyle f''(x)$ is an upwards parabola, it is positive (and thus f(x) is concave upwards) at...

Tonio

Ok but are my points of inflection correct? and there is a point where the graph concaves up and down. I plugged both answers i found for x into the orignal equation, that is how I found my points.