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 \( x= \) with output value: Check Answer

To estimate the local extrema of the function \( f(x) = 2x^3 - 36x^2 + 210x + 9 \), we can use a graph of the function. Given that the function has a local minimum at \( x = 7 \) with an output value, we can substitute \( x = 7 \) into the function to find the output value at this point. To find the local maximum, we need to find the critical points of the function by taking the derivative and setting it equal to zero. Let's first find the critical points by taking the derivative of the function and setting it equal to zero. Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dx}\left(2x^{3}-36x^{2}+210x+9\right)\) - step1: Use differentiation rules: \(\frac{d}{dx}\left(2x^{3}\right)-\frac{d}{dx}\left(36x^{2}\right)+\frac{d}{dx}\left(210x\right)+\frac{d}{dx}\left(9\right)\) - step2: Calculate: \(6x^{2}-72x+210+0\) - step3: Remove 0: \(6x^{2}-72x+210\) The derivative of the function \( f(x) = 2x^3 - 36x^2 + 210x + 9 \) is \( 6x^2 - 72x + 210 \). Now, we need to set the derivative equal to zero to find the critical points: \[ 6x^2 - 72x + 210 = 0 \] 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 are \( x = 5 \) and \( x = 7 \). Now, we can substitute these critical points back into the original function to find the local extrema. For the local minimum at \( x = 7 \): \[ f(7) = 2(7)^3 - 36(7)^2 + 210(7) + 9 \] For the local maximum at \( x = 5 \): \[ f(5) = 2(5)^3 - 36(5)^2 + 210(5) + 9 \] Let's calculate the output values for these critical points. Calculate the value by following steps: - step0: Calculate: \(2\times 7^{3}-36\times 7^{2}+210\times 7+9\) - step1: Multiply the terms: \(686-36\times 7^{2}+210\times 7+9\) - step2: Multiply the terms: \(686-1764+210\times 7+9\) - step3: Multiply the numbers: \(686-1764+1470+9\) - step4: Calculate: \(401\) Calculate or simplify the expression \( 2(5)^3 - 36(5)^2 + 210(5) + 9 \). Calculate the value by following steps: - step0: Calculate: \(2\times 5^{3}-36\times 5^{2}+210\times 5+9\) - step1: Multiply the terms: \(250-36\times 5^{2}+210\times 5+9\) - step2: Multiply the terms: \(250-900+210\times 5+9\) - step3: Multiply the numbers: \(250-900+1050+9\) - step4: Calculate: \(409\) The local minimum at \( x = 7 \) has an output value of 401, and the local maximum at \( x = 5 \) has an output value of 409. Therefore, the local minimum is at \( x = 7 \) with an output value of 401, and the local maximum is at \( x = 5 \) with an output value of 409.