The function \( f(x)=2 x^{3}-36 x^{2}+210 x+9 \) has one local minimum and one local maximum. Use a graph of the function to estimate these local extrema. This function has a local minimum at \( x=7 \) with output value: and a local maximum at \( x=\square \) with output value:

To estimate the local extrema of the function \( f(x) = 2x^3 - 36x^2 + 210x + 9 \) using a graph, we need to find the critical points of the function. Critical points occur where the derivative of the function is equal to zero or undefined. First, let's find the derivative of the function \( f(x) \) with respect to \( x \): \[ f'(x) = \frac{d}{dx}(2x^3 - 36x^2 + 210x + 9) \] Using the power rule for differentiation, we get: \[ f'(x) = 6x^2 - 72x + 210 \] Now, we need to find the critical points by setting \( f'(x) = 0 \) and solving for \( x \): \[ 6x^2 - 72x + 210 = 0 \] Let's solve this quadratic equation to find the critical points. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(6x^{2}-72x+210=0\) - step1: Factor the expression: \(6\left(x-7\right)\left(x-5\right)=0\) - step2: Divide the terms: \(\left(x-7\right)\left(x-5\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-7=0\\&x-5=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=7\\&x=5\end{align}\) - step5: Rewrite: \(x_{1}=5,x_{2}=7\) The critical points of the function \( f(x) = 2x^3 - 36x^2 + 210x + 9 \) are \( x = 5 \) and \( x = 7 \). Now, we need to determine whether these critical points correspond to local minima or maxima. To do this, we can use the second derivative test. The second derivative of the function is: \[ f''(x) = \frac{d}{dx}(6x^2 - 72x + 210) \] Using the power rule for differentiation again, we get: \[ f''(x) = 12x - 72 \] Now, we can evaluate the second derivative at the critical points to determine the nature of the extrema. 1. For \( x = 5 \): \[ f''(5) = 12(5) - 72 = 60 - 72 = -12 \] Since \( f''(5) < 0 \), the critical point \( x = 5 \) corresponds to a local maximum. 2. For \( x = 7 \): \[ f''(7) = 12(7) - 72 = 84 - 72 = 12 \] Since \( f''(7) > 0 \), the critical point \( x = 7 \) corresponds to a local minimum. Therefore, the local minimum of the function \( f(x) = 2x^3 - 36x^2 + 210x + 9 \) is at \( x = 7 \) with an output value of \( f(7) \), and the local maximum is at \( x = 5 \) with an output value of \( f(5) \).