Question

please answer the following questions about the function

$f(x)=7x^{5}-3x^{4}$.

instructions: if you are asked to find $x$- or $y$-values, enter either a number, a list of numbers
separated by commas, or none if there arent any solutions. use help (intervals) if you are asked to
find an interval or union of intervals, and enter \\{ \\} if the interval is empty.

(a) find the critical numbers of $f$, where it is increasing and decreasing, and its local extrema.

critical numbers $x=$

increasing on the interval

decreasing on the interval

local maxima $x=$

local minima $x=$

(b) find where $f$ is concave up, concave down, and has inflection points.

concave up on the interval

concave down on the interval

inflection points $x=$

Explanation:

Step1: Find the first - derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y = f(x)=7x^{5}-3x^{4}\), we have \(f^\prime(x)=35x^{4}-12x^{3}=x^{3}(35x - 12)\).
Set \(f^\prime(x)=0\), then \(x^{3}(35x - 12)=0\). Solving \(x^{3}=0\) gives \(x = 0\), and solving \(35x-12 = 0\) gives \(x=\frac{12}{35}\). So the critical numbers are \(x = 0,\frac{12}{35}\).

Step2: Determine the intervals of increase and decrease

• Choose test points:
• For the interval \((-\infty,0)\), let \(x=-1\). Then \(f^\prime(-1)=35(-1)^{4}-12(-1)^{3}=35 + 12=47>0\).
• For the interval \((0,\frac{12}{35})\), let \(x=\frac{1}{5}\). Then \(f^\prime(\frac{1}{5})=35(\frac{1}{5})^{4}-12(\frac{1}{5})^{3}=\frac{35}{625}-\frac{12}{125}=\frac{35 - 60}{625}=-\frac{25}{625}<0\).
• For the interval \((\frac{12}{35},\infty)\), let \(x = 1\). Then \(f^\prime(1)=35\times1^{4}-12\times1^{3}=35 - 12 = 23>0\).
• So \(f(x)\) is increasing on \((-\infty,0)\cup(\frac{12}{35},\infty)\) and decreasing on \((0,\frac{12}{35})\).

Step3: Find local extrema

By the first - derivative test:

• Since \(f(x)\) changes from increasing to decreasing at \(x = 0\), \(x = 0\) is a local maximum.
• Since \(f(x)\) changes from decreasing to increasing at \(x=\frac{12}{35}\), \(x=\frac{12}{35}\) is a local minimum.

Step4: Find the second - derivative

Differentiate \(f^\prime(x)=35x^{4}-12x^{3}\) using the power rule. \(f^{\prime\prime}(x)=140x^{3}-36x^{2}=4x^{2}(35x - 9)\).
Set \(f^{\prime\prime}(x)=0\), then \(4x^{2}(35x - 9)=0\). Solving \(x^{2}=0\) gives \(x = 0\), and solving \(35x-9 = 0\) gives \(x=\frac{9}{35}\).

Step5: Determine concavity

• For the interval \((-\infty,\frac{9}{35})\), let \(x = 0\). Then \(f^{\prime\prime}(0)=4\times0^{2}(35\times0 - 9)=0\). Let \(x=\frac{1}{10}\), \(f^{\prime\prime}(\frac{1}{10})=4\times(\frac{1}{10})^{2}(35\times\frac{1}{10}-9)=4\times\frac{1}{100}\times(\frac{35 - 90}{10})<0\).
• For the interval \((\frac{9}{35},\infty)\), let \(x = 1\). Then \(f^{\prime\prime}(1)=4\times1^{2}(35\times1 - 9)=4\times26>0\).
• So \(f(x)\) is concave down on \((-\infty,\frac{9}{35})\) and concave up on \((\frac{9}{35},\infty)\).
• The inflection point occurs at \(x=\frac{9}{35}\) (since the concavity changes at \(x=\frac{9}{35}\), and \(x = 0\) is not an inflection point as the concavity does not change across \(x = 0\) when considering the second - derivative test for concavity).

Answer:

• Critical numbers \(x=0,\frac{12}{35}\)
• Increasing on the interval \((-\infty,0)\cup(\frac{12}{35},\infty)\)
• Decreasing on the interval \((0,\frac{12}{35})\)
• Local maxima \(x = 0\)
• Local minima \(x=\frac{12}{35}\)
• Concave up on the interval \((\frac{9}{35},\infty)\)
• Concave down on the interval \((-\infty,\frac{9}{35})\)
• Inflection points \(x=\frac{9}{35}\)