The behavior of the graph of a polynomial function to the far left or the far right is called its_______ behavior, which depends upon the_______ term.

toofehblf
2022-02-01
Answered

The behavior of the graph of a polynomial function to the far left or the far right is called its_______ behavior, which depends upon the_______ term.

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dicky23628u6a

Answered 2022-02-02
Author has **12** answers

Explanation

Consider the polynomial function

$f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\dots +{a}_{1}x+{a}_{0}({a}_{n}\ne 0)$

In it$a}_{n$ is called the leading term.

The leading coefficient test:

Consider the polynomial function,$f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\dots +{a}_{1}x+{a}_{0}({a}_{n}\ne 0)$

That is leading coefficient is an. then,

For n is odd

$\left(a\right){a}_{n}>0$ , the graph falls to the left and rises to the right.

$\left(a\right){a}_{n}<0$ , the graph rises to the left and falls to the right

For n is even

$\left(a\right){a}_{n}>0$ , the graph rises to the left and rises to the right.

$\left(a\right){a}_{n}<0$ , the graph falls to the left and falls to the right.

Step 2

That is, Odd-degree polynomial function have graphs with opposite behavior at each end while even-degree polynomial shows same behavior at each end.

Therefore, from the leading coefficient test the end behavior is governed with the help of the leading coefficients.

Hence, the behavior of the graph of a polynomial function to the far left or the far right is called its end behavior, which depends upon the leading term.

Consider the polynomial function

In it

The leading coefficient test:

Consider the polynomial function,

That is leading coefficient is an. then,

For n is odd

For n is even

Step 2

That is, Odd-degree polynomial function have graphs with opposite behavior at each end while even-degree polynomial shows same behavior at each end.

Therefore, from the leading coefficient test the end behavior is governed with the help of the leading coefficients.

Hence, the behavior of the graph of a polynomial function to the far left or the far right is called its end behavior, which depends upon the leading term.

asked 2022-06-16

If $f(x)=\frac{\mathrm{cos}x+5\mathrm{cos}3x+\mathrm{cos}5x}{\mathrm{cos}6x+6\mathrm{cos}4x+15\mathrm{cos}2x+10}$ then find the value of $f(0)+{f}^{\prime}(0)+{f}^{\u2033}(0)$

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Find the sum of the squares of the solutions to $\sqrt{1-\mathrm{cos}x}+\sqrt{1+\mathrm{cos}x}=\sqrt{3},$, where $-\pi <x<\pi .$

Attempt: Squaring both sides gives

$1-\mathrm{cos}x+2\sqrt{(1-\mathrm{cos}x)(1+\mathrm{cos}x)}+1+\mathrm{cos}x=3,$

which simplifies to

$2\sqrt{1-{\mathrm{cos}}^{2}x}=3-1\u27fa\sqrt{{\mathrm{sin}}^{2}x}=1\u27fa|\mathrm{sin}x|=1.$

Attempt: Squaring both sides gives

$1-\mathrm{cos}x+2\sqrt{(1-\mathrm{cos}x)(1+\mathrm{cos}x)}+1+\mathrm{cos}x=3,$

which simplifies to

$2\sqrt{1-{\mathrm{cos}}^{2}x}=3-1\u27fa\sqrt{{\mathrm{sin}}^{2}x}=1\u27fa|\mathrm{sin}x|=1.$

asked 2022-01-16

Showing $\frac{d\theta}{d\mathrm{tan}\theta}=\frac{1}{1+{\mathrm{tan}}^{2}\theta}$

I suppose that

$\frac{d\theta}{d\mathrm{tan}\theta}=\frac{d\mathrm{arctan}x}{dx}=\frac{1}{1+{x}^{2}}=\frac{1}{1+{\mathrm{tan}}^{2}\theta}$

So is

$\frac{d\theta}{d\mathrm{tan}\theta}=\frac{1}{1+{\mathrm{tan}}^{2}\theta}$

correct? And

$\frac{d\theta}{d\mathrm{tan}\frac{\theta}{2}}=\frac{2}{1+{\mathrm{tan}}^{2}\frac{\theta}{2}}$ ?

I suppose that

So is

correct? And

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Evaluate

$\int \frac{dx}{\mathrm{sin}x-\mathrm{cos}x}$

I know it can be done by Weierstrass substitution. But I am looking for new/simple approach. For example I tried:

$\int \frac{1}{\mathrm{sin}x-\mathrm{cos}x}\cdot \frac{\mathrm{sin}x+\mathrm{cos}x}{\mathrm{sin}x+\mathrm{cos}x}dx=\int \frac{\mathrm{sin}x+\mathrm{cos}x}{-\mathrm{cos}2x}dx$

but I can't continue from here.

I know it can be done by Weierstrass substitution. But I am looking for new/simple approach. For example I tried:

but I can't continue from here.

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Factor. ${x}^{5}+x+1$

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Please help me prove the identity:

${\mathrm{cos}}^{2}a-{\mathrm{cos}}^{4}a+{\mathrm{sin}}^{4}a=\frac{1}{2}-\frac{1}{2}\mathrm{cos}2a$

asked 2022-07-18

Prove that $\overrightarrow{u}$ and $\overrightarrow{v}$ are orthogonal vectors

The vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are given in terms of the basis vectors $\overrightarrow{a}$ , $\overrightarrow{b}$ and $\overrightarrow{c}$ as follows:

$\overrightarrow{u}=3\overrightarrow{a}+3\overrightarrow{b}-\overrightarrow{c}$

$\overrightarrow{v}=\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}$

I've tried $\overrightarrow{u}.\overrightarrow{v}$ to see if their dot product equals to 0, but it does not. Am I missing something?

It was given that $\overrightarrow{a},\overrightarrow{b},$ and $\overrightarrow{c}$ form a basis in ${R}^{3}$.

It was also given that:

$|\overrightarrow{a}|=1,|\overrightarrow{b}|=2,|\overrightarrow{c}|=3$

The vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are given in terms of the basis vectors $\overrightarrow{a}$ , $\overrightarrow{b}$ and $\overrightarrow{c}$ as follows:

$\overrightarrow{u}=3\overrightarrow{a}+3\overrightarrow{b}-\overrightarrow{c}$

$\overrightarrow{v}=\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}$

I've tried $\overrightarrow{u}.\overrightarrow{v}$ to see if their dot product equals to 0, but it does not. Am I missing something?

It was given that $\overrightarrow{a},\overrightarrow{b},$ and $\overrightarrow{c}$ form a basis in ${R}^{3}$.

It was also given that:

$|\overrightarrow{a}|=1,|\overrightarrow{b}|=2,|\overrightarrow{c}|=3$