Find the Inverse h(x)=-(9x)/4-1/3

$h (x) = - \frac{9 x}{4} - \frac{1}{3}$

Step 1

Write $h (x) = - \frac{9 x}{4} - \frac{1}{3}$ as an equation.

$y = - \frac{9 x}{4} - \frac{1}{3}$

Step 2

Interchange the variables.

$x = - \frac{9 y}{4} - \frac{1}{3}$

Step 3

Solve for $y$ .

Rewrite the equation as $- \frac{9 y}{4} - \frac{1}{3} = x$ .

$- \frac{9 y}{4} - \frac{1}{3} = x$

Add $\frac{1}{3}$ to both sides of the equation.

$- \frac{9 y}{4} = x + \frac{1}{3}$

Multiply both sides of the equation by $- \frac{4}{9}$ .

$- \frac{4}{9} (- \frac{9 y}{4}) = - \frac{4}{9} (x + \frac{1}{3})$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{4}{9} (- \frac{9 y}{4})$ .

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{4}{9}$ into the numerator.

$\frac{- 4}{9} (- \frac{9 y}{4}) = - \frac{4}{9} (x + \frac{1}{3})$

Move the leading negative in $- \frac{9 y}{4}$ into the numerator.

$\frac{- 4}{9} \cdot \frac{- 9 y}{4} = - \frac{4}{9} (x + \frac{1}{3})$

Factor $4$ out of $- 4$ .

$\frac{4 (- 1)}{9} \cdot \frac{- 9 y}{4} = - \frac{4}{9} (x + \frac{1}{3})$

Cancel the common factor.

$\frac{4 \cdot - 1}{9} \cdot \frac{- 9 y}{4} = - \frac{4}{9} (x + \frac{1}{3})$

Rewrite the expression.

$\frac{- 1}{9} (- 9 y) = - \frac{4}{9} (x + \frac{1}{3})$

Cancel the common factor of $9$ .

Factor $9$ out of $- 9 y$ .

$\frac{- 1}{9} (9 (- y)) = - \frac{4}{9} (x + \frac{1}{3})$

Cancel the common factor.

$\frac{- 1}{9} (9 (- y)) = - \frac{4}{9} (x + \frac{1}{3})$

Rewrite the expression.

$- - y = - \frac{4}{9} (x + \frac{1}{3})$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{4}{9} (x + \frac{1}{3})$

Multiply $y$ by $1$ .

$y = - \frac{4}{9} (x + \frac{1}{3})$

Simplify the right side.

Simplify $- \frac{4}{9} (x + \frac{1}{3})$ .

Apply the distributive property.

$y = - \frac{4}{9} x - \frac{4}{9} \cdot \frac{1}{3}$

Combine $x$ and $\frac{4}{9}$ .

$y = - \frac{x \cdot 4}{9} - \frac{4}{9} \cdot \frac{1}{3}$

Multiply $- \frac{4}{9} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{3}$ by $\frac{4}{9}$ .

$y = - \frac{x \cdot 4}{9} - \frac{4}{3 \cdot 9}$

Multiply $3$ by $9$ .

$y = - \frac{x \cdot 4}{9} - \frac{4}{27}$

Move $4$ to the left of $x$ .

$y = - \frac{4 x}{9} - \frac{4}{27}$

Step 4

Replace $y$ with $h^{- 1} (x)$ to show the final answer.

$h^{- 1} (x) = - \frac{4 x}{9} - \frac{4}{27}$

Step 5

Verify if $h^{- 1} (x) = - \frac{4 x}{9} - \frac{4}{27}$ is the inverse of $h (x) = - \frac{9 x}{4} - \frac{1}{3}$ .

To verify the inverse, check if $h^{- 1} (h (x)) = x$ and $h (h^{- 1} (x)) = x$ .

Evaluate $h^{- 1} (h (x))$ .

Set up the composite result function.

$h^{- 1} (h (x))$

Evaluate $h^{- 1} (- \frac{9 x}{4} - \frac{1}{3})$ by substituting in the value of $h$ into $h^{- 1}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{4 (- \frac{9 x}{4} - \frac{1}{3})}{9} - \frac{4}{27}$

Simplify each term.

Simplify the numerator.

Factor $- 1$ out of $- \frac{9 x}{4} - \frac{1}{3}$ .

Factor $- 1$ out of $- \frac{9 x}{4}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{4 (- (\frac{9 x}{4}) - \frac{1}{3})}{9} - \frac{4}{27}$

Factor $- 1$ out of $- \frac{1}{3}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{4 (- (\frac{9 x}{4}) - (\frac{1}{3}))}{9} - \frac{4}{27}$

Factor $- 1$ out of $- (\frac{9 x}{4}) - (\frac{1}{3})$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{4 (- (\frac{9 x}{4} + \frac{1}{3}))}{9} - \frac{4}{27}$

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{4 \cdot (- 1 (\frac{9 x}{4} + \frac{1}{3}))}{9} - \frac{4}{27}$

Combine exponents.

Factor out negative.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- (4 (\frac{9 x}{4} + \frac{1}{3}))}{9} - \frac{4}{27}$

Multiply $4$ by $- 1$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 (\frac{9 x}{4} + \frac{1}{3})}{9} - \frac{4}{27}$

To write $\frac{9 x}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 (\frac{9 x}{4} \cdot \frac{3}{3} + \frac{1}{3})}{9} - \frac{4}{27}$

To write $\frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 (\frac{9 x}{4} \cdot \frac{3}{3} + \frac{1}{3} \cdot \frac{4}{4})}{9} - \frac{4}{27}$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{9 x}{4}$ by $\frac{3}{3}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 (\frac{9 x \cdot 3}{4 \cdot 3} + \frac{1}{3} \cdot \frac{4}{4})}{9} - \frac{4}{27}$

Multiply $4$ by $3$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 (\frac{9 x \cdot 3}{12} + \frac{1}{3} \cdot \frac{4}{4})}{9} - \frac{4}{27}$

Multiply $\frac{1}{3}$ by $\frac{4}{4}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 (\frac{9 x \cdot 3}{12} + \frac{4}{3 \cdot 4})}{9} - \frac{4}{27}$

Multiply $3$ by $4$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 (\frac{9 x \cdot 3}{12} + \frac{4}{12})}{9} - \frac{4}{27}$

Combine the numerators over the common denominator.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 \frac{9 x \cdot 3 + 4}{12}}{9} - \frac{4}{27}$

Multiply $3$ by $9$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- 4 \frac{27 x + 4}{12}}{9} - \frac{4}{27}$

Combine $- 4$ and $\frac{27 x + 4}{12}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{\frac{- 4 (27 x + 4)}{12}}{9} - \frac{4}{27}$

Simplify the numerator.

Reduce the expression $\frac{- 4 (27 x + 4)}{12}$ by cancelling the common factors.

Factor $4$ out of $- 4 (27 x + 4)$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{\frac{4 (- (27 x + 4))}{12}}{9} - \frac{4}{27}$

Factor $4$ out of $12$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{\frac{4 (- (27 x + 4))}{4 (3)}}{9} - \frac{4}{27}$

Cancel the common factor.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{\frac{4 (- (27 x + 4))}{4 \cdot 3}}{9} - \frac{4}{27}$

Rewrite the expression.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{\frac{- (27 x + 4)}{3}}{9} - \frac{4}{27}$

Move the negative in front of the fraction.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - \frac{- \frac{27 x + 4}{3}}{9} - \frac{4}{27}$

Multiply the numerator by the reciprocal of the denominator.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = - (- \frac{27 x + 4}{3} \cdot \frac{1}{9}) - \frac{4}{27}$

Multiply $- \frac{27 x + 4}{3} \cdot \frac{1}{9}$ .

Multiply $\frac{1}{9}$ by $\frac{27 x + 4}{3}$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = \frac{27 x + 4}{9 \cdot 3} - \frac{4}{27}$

Multiply $9$ by $3$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = \frac{27 x + 4}{27} - \frac{4}{27}$

Multiply $- - \frac{27 x + 4}{27}$ .

Multiply $- 1$ by $- 1$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = 1 (\frac{27 x + 4}{27}) - \frac{4}{27}$

Multiply $\frac{27 x + 4}{27}$ by $1$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = \frac{27 x + 4}{27} - \frac{4}{27}$

Simplify terms.

Combine the numerators over the common denominator.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = \frac{27 x + 4 - 4}{27}$

Combine the opposite terms in $27 x + 4 - 4$ .

Subtract $4$ from $4$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = \frac{27 x + 0}{27}$

Add $27 x$ and $0$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = \frac{27 x}{27}$

Cancel the common factor of $27$ .

Cancel the common factor.

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = \frac{27 x}{27}$

Divide $x$ by $1$ .

$h^{- 1} (- \frac{9 x}{4} - \frac{1}{3}) = x$

Evaluate $h (h^{- 1} (x))$ .

Set up the composite result function.

$h (h^{- 1} (x))$

Evaluate $h (- \frac{4 x}{9} - \frac{4}{27})$ by substituting in the value of $h^{- 1}$ into $h$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{9 (- \frac{4 x}{9} - \frac{4}{27})}{4} - \frac{1}{3}$

Simplify each term.

Simplify the numerator.

Factor $- 1$ out of $- \frac{4 x}{9} - \frac{4}{27}$ .

Factor $- 1$ out of $- \frac{4 x}{9}$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{9 (- (\frac{4 x}{9}) - \frac{4}{27})}{4} - \frac{1}{3}$

Factor $- 1$ out of $- \frac{4}{27}$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{9 (- (\frac{4 x}{9}) - (\frac{4}{27}))}{4} - \frac{1}{3}$

Factor $- 1$ out of $- (\frac{4 x}{9}) - (\frac{4}{27})$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{9 (- (\frac{4 x}{9} + \frac{4}{27}))}{4} - \frac{1}{3}$

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{9 \cdot (- 1 (\frac{4 x}{9} + \frac{4}{27}))}{4} - \frac{1}{3}$

Combine exponents.

Factor out negative.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- (9 (\frac{4 x}{9} + \frac{4}{27}))}{4} - \frac{1}{3}$

Multiply $9$ by $- 1$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 (\frac{4 x}{9} + \frac{4}{27})}{4} - \frac{1}{3}$

To write $\frac{4 x}{9}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 (\frac{4 x}{9} \cdot \frac{3}{3} + \frac{4}{27})}{4} - \frac{1}{3}$

Write each expression with a common denominator of $27$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4 x}{9}$ by $\frac{3}{3}$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 (\frac{4 x \cdot 3}{9 \cdot 3} + \frac{4}{27})}{4} - \frac{1}{3}$

Multiply $9$ by $3$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 (\frac{4 x \cdot 3}{27} + \frac{4}{27})}{4} - \frac{1}{3}$

Combine the numerators over the common denominator.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 \frac{4 x \cdot 3 + 4}{27}}{4} - \frac{1}{3}$

Simplify the numerator.

Factor $4$ out of $4 x \cdot 3 + 4$ .

Factor $4$ out of $4 x \cdot 3$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 \frac{4 (x \cdot 3) + 4}{27}}{4} - \frac{1}{3}$

Factor $4$ out of $4$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 \frac{4 (x \cdot 3) + 4 (1)}{27}}{4} - \frac{1}{3}$

Factor $4$ out of $4 (x \cdot 3) + 4 (1)$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 \frac{4 (x \cdot 3 + 1)}{27}}{4} - \frac{1}{3}$

Move $3$ to the left of $x$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- 9 \frac{4 (3 x + 1)}{27}}{4} - \frac{1}{3}$

Combine $- 9$ and $\frac{4 (3 x + 1)}{27}$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{\frac{- 9 (4 (3 x + 1))}{27}}{4} - \frac{1}{3}$

Multiply $- 9$ by $4$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{\frac{- 36 (3 x + 1)}{27}}{4} - \frac{1}{3}$

Simplify the numerator.

Reduce the expression $\frac{- 36 (3 x + 1)}{27}$ by cancelling the common factors.

Factor $9$ out of $- 36 (3 x + 1)$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{\frac{9 (- 4 (3 x + 1))}{27}}{4} - \frac{1}{3}$

Factor $9$ out of $27$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{\frac{9 (- 4 (3 x + 1))}{9 (3)}}{4} - \frac{1}{3}$

Cancel the common factor.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{\frac{9 (- 4 (3 x + 1))}{9 \cdot 3}}{4} - \frac{1}{3}$

Rewrite the expression.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{\frac{- 4 (3 x + 1)}{3}}{4} - \frac{1}{3}$

Move the negative in front of the fraction.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- \frac{4 (3 x + 1)}{3}}{4} - \frac{1}{3}$

Multiply the numerator by the reciprocal of the denominator.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - (- \frac{4 (3 x + 1)}{3} \cdot \frac{1}{4}) - \frac{1}{3}$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{4 (3 x + 1)}{3}$ into the numerator.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - (\frac{- 4 (3 x + 1)}{3} \cdot \frac{1}{4}) - \frac{1}{3}$

Factor $4$ out of $- 4 (3 x + 1)$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = - (\frac{4 (- (3 x + 1))}{3} \cdot \frac{1}{4}) - \frac{1}{3}$

Cancel the common factor.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - (\frac{4 (- (3 x + 1))}{3} \cdot \frac{1}{4}) - \frac{1}{3}$

Rewrite the expression.

$h (- \frac{4 x}{9} - \frac{4}{27}) = - \frac{- (3 x + 1)}{3} - \frac{1}{3}$

Move the negative in front of the fraction.

$h (- \frac{4 x}{9} - \frac{4}{27}) = \frac{3 x + 1}{3} - \frac{1}{3}$

Multiply $- - \frac{3 x + 1}{3}$ .

Multiply $- 1$ by $- 1$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = 1 (\frac{3 x + 1}{3}) - \frac{1}{3}$

Multiply $\frac{3 x + 1}{3}$ by $1$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = \frac{3 x + 1}{3} - \frac{1}{3}$

Simplify terms.

Combine the numerators over the common denominator.

$h (- \frac{4 x}{9} - \frac{4}{27}) = \frac{3 x + 1 - 1}{3}$

Combine the opposite terms in $3 x + 1 - 1$ .

Subtract $1$ from $1$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = \frac{3 x + 0}{3}$

Add $3 x$ and $0$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = \frac{3 x}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$h (- \frac{4 x}{9} - \frac{4}{27}) = \frac{3 x}{3}$

Divide $x$ by $1$ .

$h (- \frac{4 x}{9} - \frac{4}{27}) = x$

Since $h^{- 1} (h (x)) = x$ and $h (h^{- 1} (x)) = x$ , then $h^{- 1} (x) = - \frac{4 x}{9} - \frac{4}{27}$ is the inverse of $h (x) = - \frac{9 x}{4} - \frac{1}{3}$ .

$h^{- 1} (x) = - \frac{4 x}{9} - \frac{4}{27}$

Find the Inverse h(x)=-(9x)/4-1/3

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