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nekit [7.7K]
3 years ago
5

For f(x) = -4x + 1, what is the output for an input of -2?

Mathematics
2 answers:
Lisa [10]3 years ago
6 0

\huge\boxed{\boxed{9}}

Start with the original function:

f(x)=-4x+1

Substitute in the known value:

f(-2)=-4(-2)+1

Multiply:

f(-2)=8+1

Add:

f(-2)=\large\boxed{9}

gayaneshka [121]3 years ago
6 0

Answer:

9

Step-by-step explanation:

In this type of equation, x represents the input. So, all you simply have to do is  replace x with -2 and then solve.

Input: -2

f(x) = -4x + 1

f(-2) = -4(-2) + 1

f(-2) = 8 + 1

f(-2) = 9

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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}

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<u>Fundamental Theorem of Calculus</u>

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<u>Integration by substitution</u>

<u />

<u />\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

<u>Substitute</u> 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

\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}

\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta = \dfrac{1}{9} \tan \theta+\text{C}

\textsf{Use the trigonometric identity}: \quad \tan \theta=\dfrac{\sin \theta}{\cos \theta}

\implies \dfrac{\sin \theta}{9 \cos \theta} +\text{C}

\textsf{Substitute back in } \sin \theta=\dfrac{x}{3}:

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\textsf{Substitute back in }3 \cos \theta=\sqrt{9-x^2}:

\implies \dfrac{x}{9\sqrt{9-x^2}} +\text{C}

Learn more about integration by substitution here:

brainly.com/question/28156101

brainly.com/question/28155016

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