Inverse of the Function <em>y = ax²</em>

Determine the inverse of the quadratic function \(h(x) = 3x^{2}\) and sketch both graphs on the same system of axes.

Determine the inverse of the given function \(h(x)\)

• Interchange \(x\) and \(y\) in the equation.
• Make \(y\) the subject of the new equation.

\begin{align*} \text{Let } y & = 3x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = 3y^{2} \\ \cfrac{x}{3} & = y^{2} \\ \therefore y & = ±\sqrt{\cfrac{x}{3}} \qquad (x \geq 0) \end{align*}

Sketch the graphs on the same system of axes

Notice that the inverse does not pass the vertical line test and therefore is not a function.

Determine the inverse of \(q(x) = 7x^{2}\) and sketch both graphs on the same system of axes. Restrict the domain of \(q\) so that the inverse is a function.

Examine the function and determine the inverse

Determine the inverse of the function:

\begin{align*} \text{Let } y & = 7x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = 7y^{2} \\ \cfrac{x}{7} & = y^{2} \\ \therefore y & = ±\sqrt{\cfrac{x}{7}} \qquad (x \geq 0) \end{align*}

Sketch both graphs on the same system of axes

Determine the restriction on the domain

Option \(\text{1}\): Restrict the domain of \(q\) to \(x \ge 0\) so that the inverse will also be a function \(( q^{-1} )\). The restriction \(x \ge 0\) on the domain of \(q\) will restrict the range of \(q^{-1}\) such that \(y \ge 0\).

\begin{align*} q: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \\ q^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \end{align*}

or

Option \(\text{2}\): Restrict the domain of \(q\) to \(x \le 0\) so that the inverse will also be a function \(( q^{-1} )\). The restriction \(x \le 0\) on the domain of \(q\) will restrict the range of \(q^{-1}\) such that \(y \le 0\).

\begin{align*} q: \qquad & \text{domain } x \le 0 \quad \text{range } y \ge 0 \\ q^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \le 0 \end{align*}

1. Determine the inverse of \(f(x) = -x^{2}\).
2. Sketch both graphs on the same system of axes.
3. Restrict the domain of \(f\) so that its inverse is a function.

Determine the inverse of the function

\begin{align*} \text{Let } y & = -x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = -y^{2} \\ -x & = y^{2} \\ y & = ±\sqrt{-x} \qquad (x \leq 0) \end{align*}

Note: \(\sqrt{-x}\) is only defined if \(x \le 0\).

Sketch both graphs on the same system of axes

The inverse does not pass the vertical line test and is not a function.

Determine the restriction on the domain

• If \(f(x) = -x^{2}, \text{ for } x \le 0\): \begin{align*} f: \qquad & \text{domain } x \le 0 \quad \text{range } y \le 0 \\ f^{-1}: \qquad & \text{domain } x \le 0 \quad \text{range } y \le 0 \end{align*}
• If \(f(x) = -x^{2}, \text{ for } x \ge 0\): \begin{align*} f: \qquad & \text{domain } x \ge 0 \quad \text{range } y \le 0 \\ f^{-1}: \qquad & \text{domain } x \le 0 \quad \text{range } y \ge 0 \end{align*}

Given: \(h(x) = 2x^{2}, \quad x \ge 0\)

1. Determine the inverse, \(h^{-1}\).
2. Find the point where \(h\) and \(h^{-1}\) intersect.
3. Sketch \(h\) and \(h^{-1}\) on the same set of axes.
4. Use the sketch to determine if \(h\) and \(h^{-1}\) are increasing or decreasing functions.
5. Calculate the average gradient of \(h\) between the two points of intersection.

Determine the inverse of the function

\begin{align*} \text{Let } y & = 2x^{2} \qquad (x \ge 0) \\ \text{Interchange } x \text{ and } y: \quad x & = 2y^{2} \qquad (y \ge 0) \\ \cfrac{x}{2} & = y^{2} \\ y & = \sqrt{\cfrac{x}{2}} \qquad (x \ge 0, y \geq 0) \\ & \\ \therefore h^{-1}(x) & = \sqrt{\cfrac{x}{2}} \qquad (x \geq 0) \end{align*}

Determine the point of intersection

\begin{align*} 2x^{2} & = \sqrt{\cfrac{x}{2}} \\ ( 2x^{2} )^{2} & = ( \sqrt{\cfrac{x}{2}} )^{2} \\ 4x^{4} & = \cfrac{x}{2} \\ 8x^{4} & = x \\ 8x^{4} - x & = 0 \\ x(8x^{3} - 1) & = 0 \\ \therefore x = 0 &\text{ or } \enspace 8x^{3} - 1 = 0 \\ \text{If } x=0, \quad y & = 0 \\ \text{If } 8x^{3} - 1 & = 0 \\ 8x^{3} & = 1 \\ x^{3} & = \cfrac{1}{8} \\ \therefore x & = \cfrac{1}{2} \\ \text{If } x= \cfrac{1}{2}, \quad y & = \cfrac{1}{2} \end{align*}

Therefore, this gives the points A\((0;0)\) and \(B( \cfrac{1}{2} ; \cfrac{1}{2})\).

Sketch both graphs on the same system of axes

Examine the graphs

From the graphs, we see that both \(h\) and \(h^{-1}\) pass the vertical line test and therefore are functions.

\begin{align*} h: \quad & \text{as } x \text{ increases. } y \text{ also increases. } \text{ therefore } h \text{ is an increasing function. } \\ h^{-1}: \quad & \text{as } x \text{ increases. } y \text{ also increases. } \text{ therefore } h^{-1} \text{ is an increasing function. } \end{align*}

Calculate the average gradient

Calculate the average gradient of \(h\) between the points A\((0;0)\) and \(B( \cfrac{1}{2} ; \cfrac{1}{2})\).

\begin{align*} \text{Average gradient: } &= \cfrac{y_{B} - y_{A}}{x_{B} - x_{A}} \\ &= \cfrac{ \cfrac{1}{2} - 0 }{\cfrac{1}{2} - 0 } \\ &= 1 \end{align*}

Note: this is also the average gradient of \(h^{-1}\) between the points \(A\) and \(B\).