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2 years ago

I need help i'm struggling with this work

Mathematics

2 answers:

valentinak56 [21]2 years ago

6 0

Answer:

$y=0.5x+2$

Step-by-step explanation:

Finding the inverse of a function is essentially finding the complete opposite of a given function. An easy trick can be used to find the inverse of a function. This trick is essentially thinking of the evaluator (f(x)) as another variable. Solve the function for (x) in terms of (f(x)); use inverse operations to do so. After completing this process, rewrite the function such that it is in the inverse function notation; switch the position of the (f(x)) and the (x), then indicate that it is an inverse function.

$f(x)=2x-4$

Inverse operations,

$f(x)=2x-4$

$f(x)+4=2x$

$\frac{f(x)+4}{2}=x$

Simplify,

$0.5f(x)+2=x$

Rewrite in inverse function notation,

$f^-^1(x)=0.5x+2$

$y=0.5x+2$

Mamont248 [21]2 years ago

3 0

I believe the correct answer is C I could be wrong tho

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2264253

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}$

Trigonometric substitution:

$\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}$

then,

$\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}$

So the integral $\mathsf{(ii)}$ becomes

$\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}$

Integrate $\mathsf{(ii)}$ by parts:

$\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}$

$\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}$

Substitute back for the variable x, and you get

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}$

I hope this helps. =)

Tags: integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus

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2 years ago

20 points! I need help! I will also be doing a 50 point giveaway later!

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Step-by-step explanation:

I hope this helped :)

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