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Natnat's question at Yahoo! Answers regarding graphing and derivatives

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MarkFL

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Feb 24, 2012
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Here is the question:

Calculus help???! 10 points for quickest answer?


I'm trying to answer this question and I'm having a bit of a struggle

Consider the polynomial function f(x)= -5x^4+4x

a) Determine the x and y intercepts

y intercept: (at 0) = 0

x intercept (at 0) = 4/5

b) Determine the coordinates of the critical points

Critical points are 3/5 and 0.

c) Identify the nature of critical points

x<0 it's positivie
0<x<3/5 it's positive
x<3/5 it's negative

d) Determine the coordinates of the points of inflection

I got (2/5, 0.128) and (0,0)

e) Graph the function

Having trouble doing this..??? Help!

f) State the intervals of increase and decrease

g) State the intervals of concavity

x<0 concaves down
0<x<2/5 concaves up
x<2/5 concaves down

PLEASE show me how to sketch it..

Not sure if anyone is willing to do this but it's worth a try lol!

Thanks :)
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello Natnat,

We are given the function:

\(\displaystyle f(x)=-5x^4+4x\)

a) Determine the x and y intercepts

To find the $x$-intercept(s), we equate $y=f(x)$ to zero, and solve for $x$:

\(\displaystyle f(x)=-5x^4+4x=-x\left(5x^3-4 \right)=0\)

Equating each factor to zero (applying the zero-factor property, we find:

\(\displaystyle x=0\)

\(\displaystyle 5x^3-4=0\implies x=\sqrt[3]{\frac{4}{5}}\)

Hence, the $x$-intercepts are:

\(\displaystyle (0,0),\,\left(\sqrt[3]{\frac{4}{5}},0 \right)\)

To find the $y$-intercept, we equate $x$ to zero, and solve for $y$:

\(\displaystyle y=-5(0)^4+4(0)=0\)

Thus, the $y$-intercept is:

$(0,0)$

b) Determine the coordinates of the critical points

To find the critical values, we need to equate the first derivative of the function to zero:

\(\displaystyle f'(x)=-20x^3+4=-4\left(5x^3-1 \right)=0\)

Thus the critical value is:

\(\displaystyle x=\sqrt[3]{\frac{1}{5}}=\frac{1}{\sqrt[3]{5}}\)

And so the coordinates of the critical point is:

\(\displaystyle \left(\frac{1}{\sqrt[3]{5}},f\left(\frac{1}{\sqrt[3]{5}} \right) \right)=\left(\frac{1}{\sqrt[3]{5}},\frac{3}{\sqrt[3]{5}} \right)\)

c) Identify the nature of critical points

Using the second derivative test (since we will need the second derivative later to discuss concavity anyway), we first need to compute the second derivative:

\(\displaystyle f''(x)=-60x^2\)

We can see that for any non-zero value of $x$, the second derivative is negative, and so the critical point we found in part b) must be a maximum, and since there is only one turning or critical point, we may conclude that this critical point is the global maximum.

d) Determine the coordinates of the points of inflection

Equating the second derivative to zero, we find:

\(\displaystyle f''(x)=-60x^2=0\)

We find that $x$ is a root of even multiplicity, and so the second derivative will not change sign across this repeated root, hence we conclude that this function has no inflection points.

e) Graph the function

Let's wait until the end to do this once we have completed the analysis of the function.

f) State the intervals of increase and decrease

From part b) we found the critical value:

\(\displaystyle x=\frac{1}{\sqrt[3]{5}}\)

This one critical point divides the domain of all reals into two sub-intervals. Because we have already determined the extremum assocaited with this critical value is the global maximum, we may conclude:

\(\displaystyle \left(-\infty,\frac{1}{\sqrt[3]{5}} \right)\) $f(x)$ is increasing.

\(\displaystyle \left(\frac{1}{\sqrt[3]{5}},\infty \right)\) $f(x)$ is decreasing.

g) State the intervals of concavity

We have already determined that the function's second derivative is negative everywhere except for $x=0$. Hence the function is concave down on:

\(\displaystyle (-\infty,0)\,\cup\,(0,\infty)\)

Okay, now let turn our attention to sketching the graph of the function.

From part a) we found the intercepts, for a total of 2 points to plot:

\(\displaystyle (0,0),\,\left(\sqrt[3]{\frac{4}{5}},0 \right)\approx(0,0),(0.93,0)\)

From part b) we found the critical point:

\(\displaystyle \left(\frac{1}{\sqrt[3]{5}},\frac{3}{\sqrt[3]{5}} \right)\approx(0.58,1.75)\)

Now we know the function is increasing everywhere to the left of this critical and decreasing everywhere to the right of it. We also know the function is concave down everywhere except at the origin. Hence, putting all of this together, we obtain a graph closely resembling the following function plot:

natnat.jpg
 
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MarkFL

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Feb 24, 2012
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To the OP,

I am sorry, but Yahoo! Answers will not allow me to answer your question on how to access the graph in the comment field there for some reason. As a guest you should still be able to click on the attachment link to see the graph.

If that doesn't work for you try this link:

y=4x-5x^4 for x=-1 to 1 - Wolfram|Alpha