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4 years ago

The inverse of f(x) is a function

Mathematics

1 answer:

natima [27]4 years ago

8 0

Answer:

Step-by-step explanation:

Not always. If, however, the given function passes the "horizontal line test," then the given function has an inverse which is also a function. The horizontal line test consists of drawing a horizontal line through the given graph; if the line intersects the graph in only one place, it "passes," and the given function has an inverse function.

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4(x + 1) − 7x = −3 (x − 1) + 1

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The apartment complex has 20 apartment per buildings 4 apartment are 3 bedroom apartment, 7 are 2 bedroom unit,and 9 apartment are 1 bedroom units.if the apartment complex builds 50 buildings how many 3 bedroom units would they have?

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4 years ago

What’s 93/108 in simplest form

mafiozo [28]

31/36 is the correct answer

4 0

3 years ago

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Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^

serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

$\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'$

$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

(5) Use implicit differentiation here.

$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

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3 years ago

What is (-36)^5/2 please help!!

Allisa [31]

Answer:

-7776

is the answer

that's the answer dude

4 0

3 years ago

Read 2 more answers

What is the solution to -x2 + 5x - 3

CaHeK987 [17]

The solution to the equation is $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Explanation:

Given that the equation is $-x^2+5x-3=0$

We need to determine the solution of the equation.

The solution of the equation can be determined using the quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Hence, from the equation, we have,

$a=-1,\:b=5,\:c=-3$

Substituting these values in the quadratic formula, we get,

$x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$

Simplifying, we get,

$x=\frac{-5\pm \sqrt{25-12}}{-2}$

Simplifying the terms within the root, we get,

$x=\frac{-5\pm \sqrt{13}}{-2}$

Thus, the roots of the equation are $x=\frac{-5+ \sqrt{13}}{-2}$ and $x=\frac{-5- \sqrt{13}}{-2}$

Taking out the negative sign, we get,

$x=\frac{-(5- \sqrt{13})}{-2}$ and $x=\frac{-(5+ \sqrt{13})}{-2}$

Cancelling the negative sign, we get,

$x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Thus, the solutions of the equation are $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

7 0

4 years ago