Math / Algebra

QuestionFind the inverse of the function. $f(x)=9-5 x^{3}$
Hint: The cube root is the same as an exponent of $1 / 3$ , so for $\sqrt[3]{19 x}$ , you could type in $\left(19^{*} x\right)^{\wedge}(1 / 3)$ . Remember your parentheses!

Studdy Solution

STEP 1

1. The function given is $f(x) = 9 - 5x^3$ .
2. To find the inverse function, we need to solve for $x$ in terms of $y$ (where $y = f(x)$ ).
3. The inverse function, denoted as $f^{-1}(y)$ , will be expressed in terms of $y$ .

STEP 2

1. Set $f(x) = y$ and solve for $x$ .
2. Express $x$ in terms of $y$ .
3. Write the inverse function $f^{-1}(y)$ .

STEP 3

Set $f(x) = y$ .
$y = 9 - 5x^3$

STEP 4

Isolate the term involving $x$ .
$y - 9 = -5x^3$

STEP 5

Divide both sides by $-5$ to solve for $x^3$ .
$x^3 = \frac{9 - y}{5}$

STEP 6

Take the cube root of both sides to solve for $x$ .
$x = \left( \frac{9 - y}{5} \right)^{1/3}$

STEP 7

Write the inverse function $f^{-1}(y)$ .
$f^{-1}(y) = \left( \frac{9 - y}{5} \right)^{1/3}$
The inverse function of $f(x) = 9 - 5x^3$ is:
$f^{-1}(y) = \left( \frac{9 - y}{5} \right)^{1/3}$

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QuestionFind the inverse of the function. $f(x)=9-5 x^{3}$
Hint: The cube root is the same as an exponent of $1 / 3$ , so for $\sqrt[3]{19 x}$ , you could type in $\left(19^{*} x\right)^{\wedge}(1 / 3)$ . Remember your parentheses!

Studdy Solution

STEP 1

1. The function given is $f(x) = 9 - 5x^3$ .
2. To find the inverse function, we need to solve for $x$ in terms of $y$ (where $y = f(x)$ ).
3. The inverse function, denoted as $f^{-1}(y)$ , will be expressed in terms of $y$ .

STEP 2

1. Set $f(x) = y$ and solve for $x$ .
2. Express $x$ in terms of $y$ .
3. Write the inverse function $f^{-1}(y)$ .

STEP 3

Set $f(x) = y$ .
$y = 9 - 5x^3$

STEP 4

Isolate the term involving $x$ .
$y - 9 = -5x^3$

STEP 5

Divide both sides by $-5$ to solve for $x^3$ .
$x^3 = \frac{9 - y}{5}$

STEP 6

Take the cube root of both sides to solve for $x$ .
$x = \left( \frac{9 - y}{5} \right)^{1/3}$

STEP 7

Write the inverse function $f^{-1}(y)$ .
$f^{-1}(y) = \left( \frac{9 - y}{5} \right)^{1/3}$
The inverse function of $f(x) = 9 - 5x^3$ is:
$f^{-1}(y) = \left( \frac{9 - y}{5} \right)^{1/3}$

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QuestionFind the inverse of the function. f(x)=9−5x3f(x)=9-5 x^{3}f(x)=9−5x3Hint: The cube root is the same as an exponent of 1/31 / 31/3, so for 19x3\sqrt[3]{19 x}319x​, you could type in (19∗x)∧(1/3)\left(19^{*} x\right)^{\wedge}(1 / 3)(19∗x)∧(1/3). Remember your parentheses!

Studdy Solution

STEP 1

1. The function given is f(x)=9−5x3 f(x) = 9 - 5x^3 f(x)=9−5x3.2. To find the inverse function, we need to solve for x x x in terms of y y y (where y=f(x) y = f(x) y=f(x)).3. The inverse function, denoted as f−1(y) f^{-1}(y) f−1(y), will be expressed in terms of y y y.

STEP 2

1. Set f(x)=y f(x) = y f(x)=y and solve for x x x.2. Express x x x in terms of y y y.3. Write the inverse function f−1(y) f^{-1}(y) f−1(y).

STEP 3

Set f(x)=y f(x) = y f(x)=y.y=9−5x3 y = 9 - 5x^3 y=9−5x3

STEP 4

Isolate the term involving x x x.y−9=−5x3 y - 9 = -5x^3 y−9=−5x3

STEP 5

Divide both sides by −5 -5 −5 to solve for x3 x^3 x3.x3=9−y5 x^3 = \frac{9 - y}{5} x3=59−y​

STEP 6

Take the cube root of both sides to solve for x x x.x=(9−y5)1/3 x = \left( \frac{9 - y}{5} \right)^{1/3} x=(59−y​)1/3

STEP 7

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QuestionFind the inverse of the function. $f(x)=9-5 x^{3}$
Hint: The cube root is the same as an exponent of $1 / 3$ , so for $\sqrt[3]{19 x}$ , you could type in $\left(19^{*} x\right)^{\wedge}(1 / 3)$ . Remember your parentheses!

1. The function given is $f(x) = 9 - 5x^3$ .
2. To find the inverse function, we need to solve for $x$ in terms of $y$ (where $y = f(x)$ ).
3. The inverse function, denoted as $f^{-1}(y)$ , will be expressed in terms of $y$ .

1. Set $f(x) = y$ and solve for $x$ .
2. Express $x$ in terms of $y$ .
3. Write the inverse function $f^{-1}(y)$ .

Set $f(x) = y$ .
$y = 9 - 5x^3$

Isolate the term involving $x$ .
$y - 9 = -5x^3$

Divide both sides by $-5$ to solve for $x^3$ .
$x^3 = \frac{9 - y}{5}$

Take the cube root of both sides to solve for $x$ .
$x = \left( \frac{9 - y}{5} \right)^{1/3}$