Find or approximate the point(s) at which the given function equals its average value on the given interval. \( f(x)=1-\frac{x^{2}}{a^{2}} ;[0, a] \) where \( a \) is a positive real number The point(s) at which \( f(x) \) equals its average value is/are (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)

To find the point(s) at which the function \( f(x) = 1 - \frac{x^2}{a^2} \) equals its average value on the interval \([0, a]\), we will follow these steps: 1. **Calculate the average value of the function on the interval**: The average value \( A \) of a function \( f(x) \) on the interval \([b, c]\) is given by the formula: \[ A = \frac{1}{c-b} \int_b^c f(x) \, dx \] In our case, \( b = 0 \) and \( c = a \). 2. **Set up the integral**: We need to compute: \[ A = \frac{1}{a-0} \int_0^a \left(1 - \frac{x^2}{a^2}\right) \, dx \] 3. **Evaluate the integral**: We will calculate the integral: \[ \int_0^a \left(1 - \frac{x^2}{a^2}\right) \, dx \] 4. **Set \( f(x) \) equal to the average value and solve for \( x \)**: We will find the points where \( f(x) = A \). Let's perform these calculations step by step. ### Step 1: Calculate the average value \( A \) First, we compute the integral: \[ \int_0^a \left(1 - \frac{x^2}{a^2}\right) \, dx = \int_0^a 1 \, dx - \int_0^a \frac{x^2}{a^2} \, dx \] Calculating each part: 1. \(\int_0^a 1 \, dx = a\) 2. \(\int_0^a \frac{x^2}{a^2} \, dx = \frac{1}{a^2} \cdot \frac{x^3}{3} \bigg|_0^a = \frac{1}{a^2} \cdot \frac{a^3}{3} = \frac{a}{3}\) Now, substituting back: \[ \int_0^a \left(1 - \frac{x^2}{a^2}\right) \, dx = a - \frac{a}{3} = \frac{2a}{3} \] Now, we can find the average value: \[ A = \frac{1}{a} \cdot \frac{2a}{3} = \frac{2}{3} \] ### Step 2: Set \( f(x) \) equal to \( A \) Now we set \( f(x) = A \): \[ 1 - \frac{x^2}{a^2} = \frac{2}{3} \] ### Step 3: Solve for \( x \) Rearranging the equation: \[ -\frac{x^2}{a^2} = \frac{2}{3} - 1 \] \[ -\frac{x^2}{a^2} = -\frac{1}{3} \] \[ \frac{x^2}{a^2} = \frac{1}{3} \] \[ x^2 = \frac{a^2}{3} \] \[ x = \pm \frac{a}{\sqrt{3}} \] Since we are looking for points in the interval \([0, a]\), we only take the positive solution: \[ x = \frac{a}{\sqrt{3}} \] ### Final Answer The point at which \( f(x) \) equals its average value is: \[ \frac{a}{\sqrt{3}} \]