Answer :
Step 1:
Write the function
[tex]f(x)=-x^3+6x^2-9x\text{ + 7}[/tex]Step 2
To find the extremer, first, find the first derivative and then equate the first derivative to zero
[tex]f^{\prime}(x)=-3x^2+12x\text{ - 9}[/tex]Step 3:
Equate the first derivative to zero to find the critical points
[tex]\begin{gathered} -3x^2\text{ + 12x - 9 = 0} \\ x^2\text{ - 4x + 3 = 0} \\ x^2\text{ - x - 3x + 3 = 0} \\ x(x\text{ - 1) -3( x - 1) = 0} \\ (x\text{ - 3)(x - 1) = 0} \\ \text{x = 3 or 1} \end{gathered}[/tex]Step 4
Next, find the extreme values
Finding the Absolute Extrema
Find all critical numbers of f within the interval [a, b]. ...
Plug in each critical number from step 1 into the function f(x).
Plug in the endpoints, a and b, into the function f(x).
The largest value is the absolute maximum, and the smallest value is the absolute minimum.
[tex]\begin{gathered} f(x)=-x^3+6x^2-9x+7 \\ f(1)=-1^3\text{ + 6 }\times1^2\text{ - 9}\times1\text{ + 7= -1 + 6 - 9 +7 = }3 \\ f(3)\text{ = }-3^3\text{ + 6 }\times3^2\text{ -9}\times3\text{ + 7} \\ f(3)\text{ = -27 + 54 - 27 + 7 = 7} \end{gathered}[/tex]minimum (1 , 3) and maximum ( 3 , 7 )