Inverses

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GRE Quantitative Reasoning › Inverses

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If $f(x)=-\frac{3x}{5}+11$ , what is its inverse function, $f^{-1}(x)$ ?

$f^{-1}(x)=-\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{5x}{3}-\frac{55}{3}$

$f^{-1}(x)=\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{3x}{5}+\frac{33}{5}$

$f^{-1}(x)=-\frac{3x}{5}-\frac{33}{5}$

Explanation

We begin by taking $f(x)=-\frac{3x}{5}+11$ and changing the to a , giving us $y=-\frac{3x}{5}+11$ .

Next, we switch all of our and , giving us $x=-\frac{3y}{5}+11$ .

Finally, we solve for by subtracting from each side, multiplying each side by , and dividing each side by , leaving us with,

$y=-\frac{5x}{3}+\frac{55}{3}$ .

Find the inverse of the following matrix, if possible.

$A=\begin{bmatrix} 0&1 \ 1&0 \end{bmatrix}$

$\begin{bmatrix} 0& 1\ 1&0 \end{bmatrix}$

$\begin{bmatrix} 0&-1 \-1&0 \end{bmatrix}$

$\begin{bmatrix} 1& 0\ 0&1 \end{bmatrix}$

The inverse does not exist.

Explanation

Write the formula to find the inverse of a matrix.

$\begin{bmatrix} a&b\ c&d \end{bmatrix}^-^1=\frac{1}{ad-bc}\begin{bmatrix} d& -b\ -c& a \end{bmatrix}$

Using the given information we are able to find the inverse matrix.

$\begin{bmatrix} 0&1 \ 1&0 \end{bmatrix}^-^1=\frac{1}{(0)(0)-(1)(1)}\begin{bmatrix} 0&-1 \-1&0 \end{bmatrix}$

$-1\begin{bmatrix} 0&-1 \ -1&0 \end{bmatrix}=\begin{bmatrix} 0& 1\ 1&0 \end{bmatrix}$

Find the inverse of the following matrix, if possible.

$A=\begin{bmatrix} 4&-4 \ -4&-4 \end{bmatrix}$

$\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

The inverse does not exist.

$\begin{bmatrix} \frac{1}{4}&-\frac{1}{4} \ -\frac{1}{4}& -\frac{1}{4} \end{bmatrix}$

$-\frac{1}{32}$

Explanation

Write the formula to find the inverse of a matrix.

$\begin{bmatrix} a&b\ c&d \end{bmatrix}^-^1=\frac{1}{ad-bc}\begin{bmatrix} d& -b\ -c& a \end{bmatrix}$

Substituting in the given matrix we are able to find the inverse matrix.

$\begin{bmatrix} 4&-4 \ -4&-4 \end{bmatrix}^-^1=\frac{1}{(4)(-4)-(-4)(-4)}\begin{bmatrix} -4&4 \ 4&4 \end{bmatrix}$

$\frac{1}{-32}\begin{bmatrix} -4&4 \ 4&4 \end{bmatrix}=\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

Find the inverse of the following matrix, if possible.

$A=\begin{bmatrix} 4&-4 \ -4&-4 \end{bmatrix}$

$\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

The inverse does not exist.

$\begin{bmatrix} \frac{1}{4}&-\frac{1}{4} \ -\frac{1}{4}& -\frac{1}{4} \end{bmatrix}$

$-\frac{1}{32}$

Explanation

Write the formula to find the inverse of a matrix.

$\begin{bmatrix} a&b\ c&d \end{bmatrix}^-^1=\frac{1}{ad-bc}\begin{bmatrix} d& -b\ -c& a \end{bmatrix}$

Substituting in the given matrix we are able to find the inverse matrix.

$\begin{bmatrix} 4&-4 \ -4&-4 \end{bmatrix}^-^1=\frac{1}{(4)(-4)-(-4)(-4)}\begin{bmatrix} -4&4 \ 4&4 \end{bmatrix}$

$\frac{1}{-32}\begin{bmatrix} -4&4 \ 4&4 \end{bmatrix}=\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

Find the Inverse of Matrix B $(B^{-1})$ where

$B=\begin{bmatrix} 2&4 \ 5&8 \end{bmatrix}$ .

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 8&-4 \ -5& 2 \end{bmatrix}$

$B^{-1}=\frac{1}{4}\begin{bmatrix} 8&-4 \ -5& 2 \end{bmatrix}$

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 2&-4 \ -5& 8 \end{bmatrix}$

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 8&4 \ 5& 2 \end{bmatrix}$

$B^{-1}=\frac{1}{4}\begin{bmatrix} 2&4 \ 5& 8 \end{bmatrix}$

Explanation

To find the inverse matrix of B use the following formula,

$B^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d&-b \ -c& a \end{bmatrix}$ .

Since the matrix B is given as,

$B=\begin{bmatrix} 2&4 \ 5&8 \end{bmatrix}$

the inverse becomes,

$B^{-1}=\frac{1}{2*8-5*4}\begin{bmatrix} 8&-4 \ -5& 2 \end{bmatrix}$

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 8&-4 \ -5& 2 \end{bmatrix}$ .

Find the inverse of the function .

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

$f^-^1(x)=\pm \sqrt{\frac{x+1}{2}}$

$f^-^1(x)=\pm \sqrt{x-1}$

Explanation

To find the inverse of , interchange the and terms and solve for .

$\frac{x-1}{2}=y^2$

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

Find the inverse of the following function.

$y=\ln, x: +: \pi$

$y^{-1}=e^{x-\pi}$

$y^{-1}=\pi-\ln(x)$

$y^{-1}=\ln(y)-\pi$

Does not exist

Explanation

To find the inverse of y, or

$y^{-1}$

first switch your variables x and y in the equation.

$x_i=\ln(y_i)+\pi$

Second, solve for the variable in the resulting equation.

$x_i-\pi=\ln,y_i$

And by setting each side of the equation as powers of base e,

$y^{-1}=e^{x-\pi}$

Which of the following is the inverse of ?

$y=e^{\frac{x}{3}}$

$y=3e^{x}$

$y=\frac{e^x}{3}$

$y=9e^{x}$

Explanation

Which of the following is the inverse of ?

To find the inverse of a function, we need to swap x and y, and then rearrange to solve for y. The inverse of a function is basically the function we get if we swap the x and y coordinates for every point on the original function.

So, to begin, we can replace the h(x) with y.

Next, swap x and y

Now, we need to get y all by itself; we can to begin by dividng the three over.

$\frac{1}{3}x=lny$

Now, recall that

And that we can rewrite any log as an exponent as follows:

So with that in mind, we can rearrange our function to get y by itself:

$\frac{1}{3}x=lny$

Becomes our final answer:

$y=e^{\frac{x}{3}}$

Find the inverse of the following function.

$y^{-1}=\sqrt[{3}]{x-1}$

$y^{-1}=1-x^{3}}$

$y^{-1}=e^{0}}$

$y^{-1}=\sqrt[3]{x+e}$

Explanation

To find the inverse of y, or

$y^{-1}$

first switch your variables x and y in the equation.

$x_{i}=y_i^3+e^0$

Second, solve for the variable in the resulting equation.

$(x_i-e^0)^\frac{1}{3}=(y_i^{3})^\frac{1}{3}$

$y_i=\sqrt[3]{x_i-e^0}$

Simplifying a number with 0 as the power, the inverse is

$y^{-1}=\sqrt[{3}]{x-1}$

Find the inverse of the following equation.

$y=\sqrt{\ln(x)+2}$ .

$y=e^x^{^{2}-2}$

$y=\pm x^2 -2$

$y=e^\pm ^{^{x^2}-2}$

Explanation

To find the inverse in this case, we need to switch our x and y variables and then solve for y.

Therefore,

$y=\sqrt{\ln(x)+2}$ becomes,

$x=\sqrt{\ln(y)+2}$

To solve for y we square both sides to get rid of the sqaure root.

$x^2=\ln(y)+2$

We then subtract 2 from both sides and take the exponenetial of each side, leaving us with the final answer.

$x^2-2=\ln(y)$

$e^{x^2-2}=e^{\ln(y)}$

$e^x^{^{2}-2}=y$

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