Answer:
The function has a minimum
The minimum value is zero
Explanations:
Given the quadratic function expressed as shown below:
[tex]f(x)=\frac{1}{2}x^2-4x+8[/tex]
We are to determine if the function has a minimum or maximum value.
Step 1: Find the critical points of the function. At turning point f'(x) = 0.
[tex]\begin{gathered} f^{\prime}(x)=\frac{2}{2}x^{2-1}-4x^{1-1} \\ f^{\prime}(x)=x-4 \end{gathered}[/tex]
Equating the function to zero to get the critical point(s)
[tex]\begin{gathered} x-4=0 \\ x=4 \end{gathered}[/tex]
Step 2: Determine the second derivative of the function:
[tex]\begin{gathered} f^{\prime}(x)=x-4 \\ f^{\doubleprime}(x)=1 \end{gathered}[/tex]
Since f''(x) > 0, hence the function given is at the minimum.
Determine the minimum value of the function by substituting x = 4 into the original function.
[tex]\begin{gathered} f(4)=\frac{1}{2}(4)^2-4(4)+8 \\ f(4)=\frac{16}{2}-16+8 \\ f(4)=8-16+8 \\ f(4)=0 \end{gathered}[/tex]
Hence the minimum value of the function is zero
