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In order to find the minimum or maximum value of the function f(x),
[tex]f\mleft(x\mright)=10x^2+x-5[/tex]First, we have to find out at which value of x the function takes it. For example:
In order to find the value of x when it takes the maximum of minimum, we are going to analyze the derivative of the function. Then we are going to be following the next step-by-step:
STEP 1: finding the derivative of the function
STEP 2: analysis of the derivative of the function.
STEP 3: minimum or maximum value of the function
We have that the derivative of the function is given by f'(x):
[tex]\begin{gathered} f\mleft(x\mright)=10x^2+x^1-5 \\ \downarrow \\ f^{\prime}(x)=2\cdot10x^{2-1}+1\cdot x^{1-1} \\ f^{\prime}(x)=20x^{2-1}+1\cdot x^0 \\ f^{\prime}(x)=20x^1+1\cdot1 \\ f^{\prime}(x)=20x^{}+1 \end{gathered}[/tex]Then, the derivative of f(x) is:
f'(x) = 20x + 1
We have that the function has a maximum or a minimum when its derivative takes a value of 0:
[tex]\begin{gathered} f^{\prime}\mleft(x\mright)=0 \\ 0=20x+1 \end{gathered}[/tex]when this happens, then, x has a value of:
[tex]\begin{gathered} 0=20x+1 \\ \downarrow\text{ taking -1 and 20 to the left side} \\ -1=20x \\ -\frac{1}{20}=x \end{gathered}[/tex]When x=-1/20, the function takes its minimum or maximum
Now, we can replace in the equation of f(x), to see what is the value of the function when x= -1/20:
[tex]\begin{gathered} f\mleft(x\mright)=10x^2+x-5 \\ \downarrow\text{ when x=}-\frac{1}{20} \\ f(-\frac{1}{20})=10(-\frac{1}{20})^2+(-\frac{1}{20})-5 \end{gathered}[/tex]Solving f(-1/20):
[tex]\begin{gathered} f(-\frac{1}{20})=10(-\frac{1}{20})^2+(-\frac{1}{20})-5 \\ \downarrow\sin ce(-\frac{1}{20})^2=\frac{1}{400} \\ =10(\frac{1}{400})-\frac{1}{20}-5 \\ =-\frac{201}{40} \end{gathered}[/tex]Then, the minimum value of the function is
[tex]f\mleft(x\mright)=\frac{-201}{40}[/tex]Answer: -201/40
