zamenjenot7k

2022-05-01

Finding a certain antiderivative

The problem says:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}+...+\frac{{b}_{m}}{{x}^{m}}$ find a function g with ${g}^{\prime}=f$..

This problem is stated in the differentiation part of the book, integration comes later.

I tried starting with a simple example and trying to find a pattern from there:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}$

Then we can try: $g(x)=-\frac{{b}_{2}}{x}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)=\frac{{b}_{2}}{{x}^{2}}=f(x)$

Then if $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}$.

I tried: $g(x)=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{{x}^{3/2}}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)-{b}_{3}\left(\frac{-\frac{3}{2}{x}^{1/2}}{{x}^{3}}\right)$

Which is not what I wanted. The denominator in the second summand is ok, but the numerator is not what I need.

But I'm stuck on what I need to look for, any hints for deriving the correct pattern would be appreciated.

The problem says:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}+...+\frac{{b}_{m}}{{x}^{m}}$ find a function g with ${g}^{\prime}=f$..

This problem is stated in the differentiation part of the book, integration comes later.

I tried starting with a simple example and trying to find a pattern from there:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}$

Then we can try: $g(x)=-\frac{{b}_{2}}{x}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)=\frac{{b}_{2}}{{x}^{2}}=f(x)$

Then if $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}$.

I tried: $g(x)=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{{x}^{3/2}}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)-{b}_{3}\left(\frac{-\frac{3}{2}{x}^{1/2}}{{x}^{3}}\right)$

Which is not what I wanted. The denominator in the second summand is ok, but the numerator is not what I need.

But I'm stuck on what I need to look for, any hints for deriving the correct pattern would be appreciated.

Ann Mathis

Beginner2022-05-02Added 11 answers

Step 1

Solve the problem for $p(x)=\frac{b}{{x}^{k}}=b{x}^{-k}$.

Then apply the trick to each summand and sum the functions you get. Since the derivative of a sum is the sum of the derivatives, you're done.

Now, if we want to find a function q(x) such that ${q}^{\prime}=p$, we may first try with a monomial, $q(x)=a{x}^{h}$. Then we know that ${q}^{\prime}(x)=ha{x}^{h-1}$ so we can take $ha=b$ and $h-1=-k$. This means $h=-k+1$ and

$a=\frac{b}{-k+1}$ which is possible provided $k\ne 1$.

Step 2

Putting together the pieces, you can take $g(x)=\frac{{b}_{2}}{(-2+1)x}+\frac{{b}_{3}}{(-3+1){x}^{2}}+\cdots +\frac{{b}_{m}}{(-m+1){x}^{m-1}}=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{2{x}^{2}}-\cdots -\frac{{b}_{m}}{(m-1){x}^{m-1}}$

Solve the problem for $p(x)=\frac{b}{{x}^{k}}=b{x}^{-k}$.

Then apply the trick to each summand and sum the functions you get. Since the derivative of a sum is the sum of the derivatives, you're done.

Now, if we want to find a function q(x) such that ${q}^{\prime}=p$, we may first try with a monomial, $q(x)=a{x}^{h}$. Then we know that ${q}^{\prime}(x)=ha{x}^{h-1}$ so we can take $ha=b$ and $h-1=-k$. This means $h=-k+1$ and

$a=\frac{b}{-k+1}$ which is possible provided $k\ne 1$.

Step 2

Putting together the pieces, you can take $g(x)=\frac{{b}_{2}}{(-2+1)x}+\frac{{b}_{3}}{(-3+1){x}^{2}}+\cdots +\frac{{b}_{m}}{(-m+1){x}^{m-1}}=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{2{x}^{2}}-\cdots -\frac{{b}_{m}}{(m-1){x}^{m-1}}$

narratz5dz

Beginner2022-05-03Added 13 answers

Step 1

$f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}+...+\frac{{b}_{m}}{{x}^{m}}$

Say $g(x)={a}_{1}(x)+{a}_{2}(x)+...+{a}_{m-1}(x)$, such that ${a}_{i}^{\prime}(x)$ gives the ${i}^{th}$ term of f(x); that is,

$a}_{i}^{\prime}(x)=\frac{{b}_{i+1}}{{x}^{i+1}$

The first term of f(x) has an ${x}^{2}$ in the denominator. You know that when we differentiate 1/x, we get ${x}^{2}$ in the denominator. So ${a}_{1}(x)=a/x,{a}_{1}^{\prime}(x)=-a/{x}^{2}={b}_{2}/{x}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a=-{b}_{2}$.

Step 2

The second term of f(x) has an ${x}^{3}$ in the denominator. This is one degree higher than ${x}^{2}$, so it makes sense to chose ${a}_{2}(x)$ having ${x}^{2}$ in the denominator, which is one degree higher than x.

${a}_{2}(x)=c/{x}^{2},{a}_{2}^{\prime}(x)=-2c/{x}^{3}={b}_{3}/{x}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}c=-{b}_{3}/2$

Similarly chose ${a}_{3}(x)$ having ${x}^{3}$ in the denominator, to obtain $a}_{3}(x)=-\frac{{b}_{4}}{3{x}^{3}$.

You might be able to see the pattern: $a}_{i}(x)=-\frac{{b}_{i+1}}{i{x}^{i}$.

$\therefore {\displaystyle g(x)=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{2{x}^{2}}-\frac{{b}_{4}}{3{x}^{3}}...-\frac{{b}_{m}}{(m-1){x}^{m-1}}+c}$

We put an arbitrary constant c because it vanishes on differentiation.

$f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}+...+\frac{{b}_{m}}{{x}^{m}}$

Say $g(x)={a}_{1}(x)+{a}_{2}(x)+...+{a}_{m-1}(x)$, such that ${a}_{i}^{\prime}(x)$ gives the ${i}^{th}$ term of f(x); that is,

$a}_{i}^{\prime}(x)=\frac{{b}_{i+1}}{{x}^{i+1}$

The first term of f(x) has an ${x}^{2}$ in the denominator. You know that when we differentiate 1/x, we get ${x}^{2}$ in the denominator. So ${a}_{1}(x)=a/x,{a}_{1}^{\prime}(x)=-a/{x}^{2}={b}_{2}/{x}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a=-{b}_{2}$.

Step 2

The second term of f(x) has an ${x}^{3}$ in the denominator. This is one degree higher than ${x}^{2}$, so it makes sense to chose ${a}_{2}(x)$ having ${x}^{2}$ in the denominator, which is one degree higher than x.

${a}_{2}(x)=c/{x}^{2},{a}_{2}^{\prime}(x)=-2c/{x}^{3}={b}_{3}/{x}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}c=-{b}_{3}/2$

Similarly chose ${a}_{3}(x)$ having ${x}^{3}$ in the denominator, to obtain $a}_{3}(x)=-\frac{{b}_{4}}{3{x}^{3}$.

You might be able to see the pattern: $a}_{i}(x)=-\frac{{b}_{i+1}}{i{x}^{i}$.

$\therefore {\displaystyle g(x)=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{2{x}^{2}}-\frac{{b}_{4}}{3{x}^{3}}...-\frac{{b}_{m}}{(m-1){x}^{m-1}}+c}$

We put an arbitrary constant c because it vanishes on differentiation.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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