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2 years ago

Enter the amplitude of the function f(x) . f(x)=3/4cos(3x)+4

Mathematics

2 answers:

Serhud [2]2 years ago

5 0

ANSWER

$\frac{3}{4}$

EXPLANATION

The given function is

$f(x) = \frac{3}{4} \cos(3x) + 4$

This function is of the form:

$y = a \cos(bx + c) + d$

where |a| is the amplitude of the function.

Since

$a = \frac{3}{4}$

$| \frac{3}{4} | = \frac{3}{4}$

Hence, the amplitude of the given cosine function is ¾.

Alisiya [41]2 years ago

4 0

Answer:

The amplitude is $\frac{3}{4}$

Step-by-step explanation:

Let f(x) be a sinusoid function that has the following format:

$f(x) = bsin(wx + s) + h$

Then b is the amplitude of the function

h is the vertical displacement of the wave

s is the phase angle

$\frac{2\pi}{w}$ is the period.

In this problem $f (x) = \frac{3}{4}cos(3x) +4$

Then $\frac{3}{4} = b$

Therefore the amplitude of the function is $\frac{3}{4}$

The amplitude is a measure of the vertical size of the sinx wave

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2 years ago

Evaluate the integral, show all steps please!

Aloiza [94]

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$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

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