Algebra 3/4

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Find the inverse of .

$f(x)=\frac{3}{2+5x}$

$f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}$

$f^{-1}(x)=\frac{3}{5x}+\frac{2}{5}$

$f^{-1}(x)=\frac{15}{x}-\frac{2}{5}$

$f^{-1}(x)=\frac{3}{5x}-10$

$f^{-1}(x)=\frac{15}{x}+\frac{2}{5}$

Explanation

To find the inverse of a function swap the variables and solve for . The function and its inverse when multiplied together, equals one. This means that the inverse undoes the function.

For this particular function the inverse is found as follows.

$f(x)=\frac{3}{2+5x}$

$y=\frac{3}{2+5x}$

First, switch the variables.

$x=\frac{3}{2+5y}$

Now, perform algebraic operations to solve for .

$x\cdot (2+5y)={3}$

$\frac{x\cdot(2+5y)}{x}=\frac{3}{x}$

$\2+5y=\frac{3}{x} \\5y=\frac{3}{x}-2 \\y=\frac{3}{5x}-\frac{2}{5}$

Therefore, the inverse is

$f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}$

Find the composition, $f\circ g$ given

$\f(x)=x^2+2x \g(x)=3x$

$f\circ g=9x^2+6x$

$f\circ g=9x^2+6$

$f\circ g=3x^2+6x$

$f\circ g=3x^2+3x$

$f\circ g=3x^2+6$

Explanation

To find the composition, $f\circ g$ given

$\f(x)=x^2+2x \g(x)=3x$

recall what a composition of two functions represents.

$f \circ g=f(g(x))$

This means that the function will replace each in the function .

Therefore, in this particular problem the composition becomes

$\f\circ g=f(g(x)) \f(g(x))=(3x)^2+2(3x) \f(g(x))=9x^2+6x$

Convert the expression from logarithmic to exponential.

$\log_4 1024=5$

$1024^\frac{1}{4}=5$

Explanation

To convert the logarithmic expression to exponential form, recall the change of base formula.

$\log_b a=c\rightarrow b^c=a$

Apply the change of base formula to this particular logarithmic expression.

$\log_4 1024=5$

$\b=4 \a=1024 \c=5$

The exponential form becomes

What is the degree equivalent for an angle that is $\frac{\pi}{2}$ radians?

$90^\circ$

$180^\circ$

$45^\circ$

$30^\circ$

$135^\circ$

Explanation

To find the degree equivalent for an angle that is $\frac{\pi}{2}$ radians first recall the unit conversion between degrees and radians.

There are 360 degrees or $2\pi$ radians in a circle.

When simplified this is,

$\pi=180^\circ$

therefore to convert from radians to degrees, simply multiply by $\frac{180^\circ}{\pi}$ .

$\frac{\pi}{2}\times \frac{180^\circ}{\pi}$

$\frac{180^\circ}{2}=90^\circ$

Find the inverse of .

$f(x)=\frac{3}{2+5x}$

$f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}$

$f^{-1}(x)=\frac{3}{5x}+\frac{2}{5}$

$f^{-1}(x)=\frac{15}{x}-\frac{2}{5}$

$f^{-1}(x)=\frac{3}{5x}-10$

$f^{-1}(x)=\frac{15}{x}+\frac{2}{5}$

Explanation

To find the inverse of a function swap the variables and solve for . The function and its inverse when multiplied together, equals one. This means that the inverse undoes the function.

For this particular function the inverse is found as follows.

$f(x)=\frac{3}{2+5x}$

$y=\frac{3}{2+5x}$

First, switch the variables.

$x=\frac{3}{2+5y}$

Now, perform algebraic operations to solve for .

$x\cdot (2+5y)={3}$

$\frac{x\cdot(2+5y)}{x}=\frac{3}{x}$

$\2+5y=\frac{3}{x} \\5y=\frac{3}{x}-2 \\y=\frac{3}{5x}-\frac{2}{5}$

Therefore, the inverse is

$f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}$

Convert the expression from logarithmic to exponential.

$\log_4 1024=5$

$1024^\frac{1}{4}=5$

Explanation

To convert the logarithmic expression to exponential form, recall the change of base formula.

$\log_b a=c\rightarrow b^c=a$

Apply the change of base formula to this particular logarithmic expression.

$\log_4 1024=5$

$\b=4 \a=1024 \c=5$

The exponential form becomes

What is the degree equivalent for an angle that is $\frac{\pi}{2}$ radians?

$90^\circ$

$180^\circ$

$45^\circ$

$30^\circ$

$135^\circ$

Explanation

To find the degree equivalent for an angle that is $\frac{\pi}{2}$ radians first recall the unit conversion between degrees and radians.

There are 360 degrees or $2\pi$ radians in a circle.

When simplified this is,

$\pi=180^\circ$

therefore to convert from radians to degrees, simply multiply by $\frac{180^\circ}{\pi}$ .

$\frac{\pi}{2}\times \frac{180^\circ}{\pi}$

$\frac{180^\circ}{2}=90^\circ$

Find the composition, $f\circ g$ given

$\f(x)=x^2+2x \g(x)=3x$

$f\circ g=9x^2+6x$

$f\circ g=9x^2+6$

$f\circ g=3x^2+6x$

$f\circ g=3x^2+3x$

$f\circ g=3x^2+6$

Explanation

To find the composition, $f\circ g$ given

$\f(x)=x^2+2x \g(x)=3x$

recall what a composition of two functions represents.

$f \circ g=f(g(x))$

This means that the function will replace each in the function .

Therefore, in this particular problem the composition becomes

$\f\circ g=f(g(x)) \f(g(x))=(3x)^2+2(3x) \f(g(x))=9x^2+6x$

Convert the expression from logarithmic to exponential.

$\log_6 216=3$

$216=6^{1/3}$

$216=3^{1/6}$

$216^{1/6}=3$

Explanation

To convert the logarithmic expression to exponential form, recall the change of base formula.

$\log_b a=c\rightarrow b^c=a$

Apply the change of base formula to this particular logarithmic expression.

$\log_6 216=3$

$\b=6 \a=216 \c=3$

The exponential form becomes

Find the composition, $f\circ g$ given

$\f(x)=x^2+2x \g(x)=3x$

$f\circ g=9x^2+6x$

$f\circ g=9x^2+6$

$f\circ g=3x^2+6x$

$f\circ g=3x^2+3x$

$f\circ g=3x^2+6$

Explanation

To find the composition, $f\circ g$ given

$\f(x)=x^2+2x \g(x)=3x$

recall what a composition of two functions represents.

$f \circ g=f(g(x))$

This means that the function will replace each in the function .

Therefore, in this particular problem the composition becomes

$\f\circ g=f(g(x)) \f(g(x))=(3x)^2+2(3x) \f(g(x))=9x^2+6x$

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