# Thread: Calculus Question: Relative Max/Min and Inflection Points?

1. ## Calculus Question: Relative Max/Min and Inflection Points?

For the function y = f(x), the solutions to the equation 0 = f'(x) are x = -3,0,2 and the solutions to the equation 0 - f"(x) are x= -1, 2. Furthermore, f"(-2) = 7, f"(1) = -9, and f"(3) = 4.

a. Where does y = f(x) have a relative maximum and where does y = f(x) have a relative minimum?
b. Where does y= f(x) have an inflection point?

Would love some help with this, I don't even know where to begin!

2. ## Re: Calculus Question: Relative Max/Min and Inflection Points?

The sign of the $f'(x_0)$ tells you if the function $f$ is increasing or decreasing in the neighborhood of $x_0$. Note that on left of the relative maximum $f$ must be an increasing function, and on the right of the relative maximum it must be a decreasing function. It is the other way around in the neighborhood of a relative minimum.

And bear in mind that what $f'$ tells you about $f$, $f''$ tells you about $f'$.

Solution to the $f''(x)=0$ is the candidate for the inflection point, and it actually is an inflection point if on the left and on the right of it $f''$ is of the opposite sign.

3. ## Re: Calculus Question: Relative Max/Min and Inflection Points?

So how do I go about answering this? Where is the starting point?

4. ## Re: Calculus Question: Relative Max/Min and Inflection Points?

First, $x=-1$ and $x=2$ are inflection points for function $f$ since $f''$ is of the opposite sign on the opposite sides of either of those two numbers.

5. ## Re: Calculus Question: Relative Max/Min and Inflection Points?

So to answer B, the inflection points are x=-1 and x=2? I do not understand how to get relative maximum and minimum since an equation was never given (or I don't know how to get an equation from the derivative values). My instructor has never presented us with a problem like this. I would appreciate being walked through it step-by-step since I have an exam over it tomorrow. If that would be at all possible, I would appreciate it.

6. ## Re: Calculus Question: Relative Max/Min and Inflection Points?

Those inflection points split up the domain (suppose $\mathb{R}$) into intervals: $\langle-\infty, -1\rangle$, $\langle -1,2\rangle$, $\langle 2, +\infty \rangle$. On each of those intervals you know the sign of $f''$ since it can change sign only in inflection points.

If you know that $f''(x)>0, \forall x\in \langle a,b \rangle$ than $f$ is convex on $\langle a,b \rangle$. Otherwise $f$ is said to be concave.

Since you know that $f''(-2)=7>0$ and $-2\in \langle-\infty, -1\rangle$ then $f''(x)>0, \forall x\in\langle-\infty, -1\rangle$, therefore $f$ is convex on $\langle-\infty, -1\rangle$. Therefore the stationary point $x=-3\in \langle -\infty,-1\rangle$ can only be relative minimum.

Since you know that $f''(1)=-9<0$ and $1\in\langle -1,2\rangle$ then $f''(x)<0, \forall x\in\langle -1,2\rangle$, therefore $f$ is concave on $\langle -1,2\rangle$. Therefore the stationary point $x=0\in\langle -1,2\rangle$ can only be relative maximum.

7. ## Re: Calculus Question: Relative Max/Min and Inflection Points?

Do the last one yourself.