Find the absolute max value of f on the interval [0,4] when f(x)= 5-6x+x^2?

Question

A) -3
B) 6
C) 5
D) -5
E) -4

Answer 1

f'(x)=2x-6->f'(x)=0,x = 3.
so the answer is or in one end of the interval or in the extrema point.
f(0)=5,f(3)=-4, f(4)=-3.
The answer is C (5)

Answer 2

5

Answer 3

The answer is C) 5
f(x) is a parabola opening upward on this interval its vertex is at (3, -4). So, the maximums would appear at the end points of the interval x = 0 and x = 4. f(0) = 5 and f(4) = -3
so the abs max = 5

Answer 4

f(x) = y = x^2-6x+5 ,Now find the 1st derivative of f(x)

f'(x) = 2x - 6 and the second derivative

f"(x) = 2 . Now put the value of x in the f'(x) .We can get
f'(0) = -6 and f'(4) = 2 ,So now because f"(x) is positive and for x = 0 , f'(x) is negative , so this function will have maximum value for x = 0

So the maximum vaue of f(x) = 5 and its absolute value = +5 .Therefore 3rd option is right

Answer 5

First, distribute the unfavourable signes in the front of the 2d & third instruments. (x^2-11x+4) + (2x^2-2x+6) + (-x^2-6x+5) Now eliminate the parenthesis and combine like variables (ie. in undemanding words upload x^2's at the same time and in undemanding words upload x's at the same time.) 2x^2 -19x +15 it truly is your anser.

Answer 6

f(x)=x^2-6x+5
x1,2= (6+-4)/2=>x1=1
x2=5
f(x) minim in x=3 ;
so f(0)=maxim =5