Find the first derivative of the function.

Step 1: Find the first derivative of the function. The given function is $y = \frac{1}{3}x^3 - x + 7$. To find the turning points, we first calculate the first derivative $\frac{dy}{dx}$. $$ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - x + 7\right) $$ $$ \frac{dy}{dx} = \frac{1}{3}(3x^2) - 1 + 0 $$ $$ \frac{dy}{dx} = x^2 - 1 $$ Step 2: Set the first derivative to zero to find the critical points. $$ x^2 - 1 = 0 $$ $$ x^2 = 1 $$ $$ x = \pm\sqrt{1} $$ So, $x = 1$ or $x = -1$. These are the x-coordinates of the turning points. Step 3: Find the y-coordinates of the turning points. Substitute the x-values back into the original function $y = \frac{1}{3}x^3 - x + 7$. For $x = 1$: $$ y = \frac{1}{3}(1)^3 - (1) + 7 $$ $$ y = \frac{1}{3} - 1 + 7 $$ $$ y = \frac{1}{3} + 6 $$ $$ y = \frac{1}{3} + \frac{18}{3} $$ $$ y = \frac{19}{3} $$ The first turning point is $\left(1, \frac{19}{3}\right)$. For $x = -1$: $$ y = \frac{1}{3}(-1)^3 - (-1) + 7 $$ $$ y = \frac{1}{3}(-1) + 1 + 7 $$ $$ y = -\frac{1}{3} + 8 $$ $$ y = -\frac{1}{3} + \frac{24}{3} $$ $$ y = \frac{23}{3} $$ The second turning point is $\left(-1, \frac{23}{3}\right)$. Step 4: Find the second derivative to classify the turning points. The first derivative is $\frac{dy}{dx} = x^2 - 1$. $$ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^2 - 1) $$ $$ \frac{d^2y}{dx^2} = 2x $$ Step 5: Classify the turning points using the second derivative test. For $x = 1$: $$ \frac{d^2y}{dx^2} = 2(1) = 2 $$ Since $\frac{d^2y}{dx^2} > 0$, the turning point $\left(1, \frac{19}{3}\right)$ is a local minimum. For $x = -1$: $$ \frac{d^2y}{dx^2} = 2(-1) = -2 $$ Since $\frac{d^2y}{dx^2} < 0$, the turning point $\left(-1, \frac{23}{3}\right)$ is a local maximum. The turning points are: Local minimum: $\boxed{\left(1, \frac{19}{3}\right)}$ Local maximum: $\boxed{\left(-1, \frac{23}{3}\right)}$ That's 2 down. 3 left today — send the next one.