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 =) and a local maximum at (x =) question help: video 1 video 2 enter a mathematical expression more...

Explanation:

Step1: Find the derivative of the function.

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

Step2: Set the derivative equal to zero to find critical points.

\[

$$\begin{align*} 6x^{2}-90x + 300&=0\\ x^{2}-15x + 50&=0\\ (x - 5)(x - 10)&=0 \end{align*}$$

\]
The critical points are \(x = 5\) and \(x=10\).

Step3: Use the second - derivative test to determine if the critical points are local maxima or minima.

Find the second - derivative \(f^{\prime\prime}(x)=12x-90\).
For \(x = 5\), \(f^{\prime\prime}(5)=12\times5-90=60 - 90=- 30<0\). So \(x = 5\) is a local maximum.
For \(x = 10\), \(f^{\prime\prime}(10)=12\times10-90=120 - 90 = 30>0\). So \(x = 10\) is a local minimum.

Answer:

Local minimum at \(x = 10\), local maximum at \(x = 5\)