A Function's Intervals of Increasing/Decreasing, Extrema and Concavity

[omwattie]

Suppose that
f(x) = (x^2 + 10)(4 - x^2).

(A) Find all critical values of f.Critical value(s) =
(B) Use interval notation to indicate where f(x) is increasing. Increasing: =
(C) Use interval notation to indicate where f(x) is decreasing. Decreasing: =
D) Find the x-coordinates of all local maxima of f. Local Maxima:=
E) Find the x-coordinates of all local minima of f. Local Minima:=
(F) Use interval notation to indicate where f(x) is concave up. Concave up:=
(G) Use interval notation to indicate where f(x) is concave down. Concave down:=

 

[MarkFL]

I've taken the liberty of retitling your thread so that it describes the nature of the problem being discussed. This helps with searches and site organization in general.

We ask that our users show what they have tried so far so we know how best to help.

You are being asked about intervals of increasing/decreasing (slope) of a given function. What do we need to be able to analyze a function's slope?

 

[omwattie]

Yes curve sketching

 

[MarkFL]

Yes curve sketching

Well, yes...answering the given questions about the function will be a good aid in sketching its graph, but let's start with part (A). We are asked to identify the function's critical values. How do we go about finding a function's critical values?

I am asking to see how much you know about what needs to be done. :)

 

[omwattie]

Sure I have no problem, well first of all you find the derivative of the originally equation, which I did but the answers is wrong.

 

[MarkFL]

Yes, to find the critical values, we first need to find the first derivative and then equate this derivative to zero, and solve for $x$ to get the critical values. Let's look at the function:

\(\displaystyle f(x)=\left(x^2+10\right)\left(4-x^2\right)\)

Now, we have a couple of choices here...we can leave the function as is and use the product/power rules, or we can expand the function and then compute the derivative using the power rule on each term.

Can you show what you did to get the derivative so I can see what you may have done incorrectly?

 

[omwattie]

I used the quotient rules which gives me
(4-x^2)(2x)-(x^2+10)(-2x)/(4-x^2)^2=28x/(4-x^2)^2
CV=-28,-2,0,2,16

I also did it another way by multiplied the two equations but the answers look weird

 

[MarkFL]

I used the quotient rules which gives me
(4-x^2)(2x)-(x^2+10)(-2x)/(4-x^2)^2=28x/(4-x^2)^2
CV=-28,-2,0,2,16

I also did it another way by multiplied the two equations but the answers look weird

In your first post, you state:

\(\displaystyle f(x)=\left(x^2+10\right)\left(4-x^2\right)\)

Is it actually supposed to be:

\(\displaystyle f(x)=\frac{x^2+10}{4-x^2}\)?

 

[omwattie]

yeah but the equation are multiplying not dividing.

 

[MarkFL]

yeah but the equation are multiplying not dividing.

In your post (#7 in this thread), where you showed your differentiation of the function, you applied the quotient rule...leading me to suspect the function is a quotient, rather than a product as you original gave. So, assuming we actually have:

\(\displaystyle f(x)=\left(x^2+10\right)\left(4-x^2\right)\)

Then, as I stated in post #6, we can wither apply the product/power rules, or expand the function and apply the power rule to each term. Can you choose one of these options and compute $f'(x)$?

 

[omwattie]

Okay what I did before I tried solving it with the quotient rule which was wrong, so I used the product rule which gives me the;
F'(x) = -4x^3-12x

- - - Updated - - -

Sorry for the confusion

 

[MarkFL]

No worries! :D

I just wanted to be certain we have the function correctly stated before proceeding.

Given:

\(\displaystyle f(x)=\left(x^2+10\right)\left(4-x^2\right)\)

Then, using the product/power rules, we find:

\(\displaystyle f'(x)=\left(x^2+10\right)(-2x)+\left(4-x^2\right)(2x)=2x\left(-x^2-10+4-x^2\right)=2x\left(-2x^2-6\right)=-4x\left(x^2+3\right)\)

This is equivalent to what you wrote, but if is factored which will make finding the roots of this derivative (the critical values) easier. So, what are the critical values?

 

[omwattie]

There is no critical value(s)

 

[MarkFL]

There is no critical value(s)

That's not correct...what we want to do here is take each factor involving $x$ in turn, equate it to zero, and solve for $x$ (for real values). So, doing this, we have:

\(\displaystyle x=0\tag{1}\)

\(\displaystyle x^2+3=0\tag{2}\)

Equation (1) is already solved for $x$, so this yields the critical value $x=0$. What critical values does equation (2) yield?

 

[omwattie]

sqrt(3),-sqrt(3)

 

[MarkFL]

sqrt(3),-sqrt(3)

Those are the roots for:

\(\displaystyle x^2-3=0\)

But, we have:

\(\displaystyle x^2+3=0\)

And this has no real roots (its roots are imaginary: $x=\pm\sqrt{3}i$). Think of the graph of:

\(\displaystyle y=x^2+3\)

It lies entirely above the $x$-axis for all of its domain, having the minimum value of $y(0)=3$.

Okay, so we have found that $f$ has one critical value, which is $x=0$. Now, knowing the factor $x^2+3$ is always positive, then to find the sign of the derivative (which relates directly to whether the function is increasing or decreasing), we need only look at the sign of the factor $-4x$.

So, can you give the intervals in which $f$ in increasing/decreasing?

I have to run for several hours, so if anyone else wants to jump in, please feel free to do so. ;)

 

[omwattie]

I figured out the where f (x) is increasing and decreasing but I don't know the concave up and concave down

 

[MarkFL]

I figured out the where f (x) is increasing and decreasing but I don't know the concave up and concave down

What do we need first before analyzing a function's concavity?

 

[omwattie]

I believe the second derivative

 

[MarkFL]

I believe the second derivative

Yes, that's correct...and what do you have for $f''$?

 

[omwattie]

I have -12x^2-12

 

[MarkFL]

Okay, that's correct:

\(\displaystyle f''(x)=-12x^2-12\)

What you want to do now, is find the (real) roots of this second derivative, so you can identify the (potential) points of inflection, and the intervals where the function is positive (concave up) and where the function is negative (concave down). What do you find?

 

[omwattie]

I found (0,-12)

 

[MarkFL]

I found (0,-12)

You want to find $(x,0)$ instead...that is, you want to find all of the roots (zeroes) of the second derivative function. So, set $f''(x)=0$, and solve for $x$...what do you find?

 

[omwattie]

(0,1,-1)

 

[MarkFL]

(0,1,-1)

No, you want to equate $f''$ to zero and solve for $x$:

\(\displaystyle -12x^2-12=0\)

Divide through by -12:

\(\displaystyle x^2+1=0\)

We should recognize that this has no real roots...and we see that the second derivative is negative for all real values of $x$, and so the given function is concave down on:

\(\displaystyle (-\infty,\infty)\)