Question

the function (f(x)=2x^{3}-45x^{2}+300x + 7) 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 =) with output value and a local maximum at (x =) with output value question help: video 1 video 2 video 3 message instructor

Explanation:

Step1: Find the derivative of the function

Given $f(x)=2x^{3}-45x^{2}+300x + 7$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=6x^{2}-90x + 300$.

Step2: Set the derivative equal to zero

Set $f'(x)=0$, so $6x^{2}-90x + 300 = 0$. Divide through by 6 to simplify: $x^{2}-15x + 50=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation: $(x - 5)(x - 10)=0$. Then $x=5$ or $x = 10$.

Step4: Use the second - derivative test

Find the second - derivative $f''(x)=12x-90$.
When $x = 5$, $f''(5)=12\times5-90=60 - 90=-30<0$, so $f(x)$ has a local maximum at $x = 5$.
When $x = 10$, $f''(10)=12\times10-90=120 - 90 = 30>0$, so $f(x)$ has a local minimum at $x = 10$.

Answer:

The function has a local minimum at $x = 10$ and a local maximum at $x = 5$.