minimum:10
output :509
maximum:5
output: 634
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
[tex]f(x)=2x^3-45x^2+300x+9[/tex]a) derivate
[tex]\begin{gathered} f(x)=2x^3-45x^2+300x+9 \\ \end{gathered}[/tex]To take the derivative of this equation, we must use the power rule
[tex]\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}[/tex]solve for x by applying the quadratic formula
[tex]ax^2+bx+c=0\rightarrow6x^2-90x+300=0[/tex][tex]\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=\frac{90\pm30}{12} \\ so \\ x_1=\frac{90+30}{12}=\frac{120}{12}=10 \\ x_2=\frac{90-30}{12}=\frac{60}{12}=5 \end{gathered}[/tex]then, let's find the output when x=5
[tex]\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}[/tex]so,infelection point is (5,634)
Step 3
Now, the output when x=10
[tex]\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}[/tex]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
[tex]\begin{gathered} f^{\prime}(x)=6x^2-90x+300 \\ f^{\prime^{\prime}}(x)=12x^{}-90 \end{gathered}[/tex]now, check if f''(x) is greater than zero
a)at x=5
[tex]\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}[/tex]it is smaller than zero, it means (5,634) is a maximum
b)at x=10
[tex]\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}[/tex]it is greater than zero, it means (10,509) is a minimum
I hope this helps you
