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MarkFL
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Here is the question:
I have posted a link there to this thread so the OP can view my work.
I have posted a link there to this thread so the OP can view my work.
Natnat's question at Yahoo Answers regarding graphing and derivatives
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In summary, we are given a polynomial function and were asked to determine its x and y intercepts, as well as its critical points, nature of critical points, points of inflection, and intervals of increase and decrease. We found that the function has two x intercepts at (0,0) and (4/5,0), its critical point is at (1/∛5, 3/∛5), and it has no points of inflection. The function is increasing on (-∞,1/∛5) and decreasing on (1/∛5, ∞), and it is concave down on (-∞,0) and (0,∞). A graph of the
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MarkFL
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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:
View attachment 1843
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:
View attachment 1843
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MarkFL
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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
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
1. How do I graph a function and its derivative on the same graph?
To graph a function and its derivative on the same graph, first plot the original function with a solid line. Then, plot the derivative function with a dashed line. Make sure to label each line with the corresponding function.
2. What is the purpose of graphing derivatives?
Graphing derivatives allows us to visualize the rate of change of a function at any given point. It also helps us identify critical points, local extrema, and concavity of a function.
3. Can I use a graphing calculator to graph derivatives?
Yes, most graphing calculators have the capability to graph derivatives. You can use the "dy/dx" or "deriv" feature to plot the derivative of a function.
4. How do I find the derivative of a function?
To find the derivative of a function, you can use the power rule, product rule, quotient rule, or chain rule depending on the form of the function. You can also use software or online tools to calculate the derivative for you.
5. What is the relationship between a function and its derivative?
The derivative of a function represents the rate of change of the function at any given point. It also tells us the slope of the tangent line to the function at that point. The derivative and the original function are related through the process of differentiation.
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