Question

This function has a local minimum at x= with output value: And a local maximum at x=With output value:

Answer 1

minimum:10

output :509

maximum:5

output: 634

Step-by-step explanation

Step 1

graph the function:

to do this, you need put values for x, and you will get a set of values for y, those formed pairs are the coordinates,

we can see, there are a minimun and a maximum

Step 2

find the minimum

To find the local minimum of any graph, you must first take the derivative of the graph equation, set it equal to zero and solve for

`f(x)=2x^3-45x^2+300x+9`

a) derivate

`\begin{gathered} f(x)=2x^3-45x^2+300x+9 \\ \end{gathered}`

To take the derivative of this equation, we must use the power rule

`\begin{gathered} f(x)=2x^3-45x^2+300x+9 \\ f^(\prime)(x)=(2\cdot3)x^((3-1))-(45\cdot2)x^(2-1)+300x^((1-1))+9 \\ f^(\prime)(x)=6x^2-90x+300 \end{gathered}`

solve for x by applying the quadratic formula

`ax^2+bx+c=0\rightarrow6x^2-90x+300=0`

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{replace} \\ x=\frac{-(-90)\pm\sqrt[]{90^2-4\cdot6\cdot300}}{2\cdot6} \\ x=\frac{90\pm\sqrt[]{8100-7200}}{12} \\ x=\frac{90\pm\sqrt[]{900}}{12} \\ x=(90\pm30)/(12) \\ so \\ x_1=(90+30)/(12)=(120)/(12)=10 \\ x_2=(90-30)/(12)=(60)/(12)=5 \end{gathered}`

then, let's find the output when x=5

`\begin{gathered} f(x)=2x^3-45x^2+300x+9 \\ f(5)=2\cdot5^3-45\cdot5^2+300\cdot5+9 \\ f(5)=250-1125+1500+9 \\ f(5)=634 \end{gathered}`

so,infelection point is (5,634)

Step 3

Now, the output when x=10

`\begin{gathered} f(x)=2x^3-45x^2+300x+9 \\ f(10)=2\cdot10^3-45\cdot10^2+300\cdot10+9 \\ f(10)=2000-4500+3000+9 \\ f(10)=509 \end{gathered}`

inflection point (10,509)

Step 3

so, at x=5 and x=10 we have two inflection points, to know if those points are minimum we need to check the second derivate of the fucntion

`\begin{gathered} f^(\prime)(x)=6x^2-90x+300 \\ f^{\prime^(\prime)}(x)=12x^{}-90 \end{gathered}`

now, check if f''(x) is greater than zero

a)at x=5

`\begin{gathered} f^{\prime^(\prime)}(x)=12x^{}-90 \\ f^{\prime^(\prime)}(5)=12\cdot5^{}-90 \\ f^{\prime^(\prime)}(5)=60-90=-30 \\ f^{\prime^(\prime)}(5)=-30 \end{gathered}`

it is smaller than zero, it means (5,634) is a maximum

b)at x=10

`\begin{gathered} f^{\prime^{}}^(\prime)(x)=12x^{}-90 \\ f^{\prime^{}\prime}(10)=12\cdot10-90 \\ f^{\prime^{}\prime}(10)=120-90 \\ f^{\prime^{}\prime}(10)=30 \end{gathered}`

it is greater than zero, it means (10,509) is a minimum

I hope this helps you