Find the absolute max value of f on the interval [0,4]?

Question

when f(x)= 5-6x+x^2
A) abs. max value= -3
B) 6
C) 5
D) -5
E) -4

Answer 1

This is a parabola that opens upward. So the vertex is the absolute min, and the absolute max would most likely be one of the endpoints of the interval. If it opened downward the vertex would be the absolute max.
Did you copy f(x) correctly?
Generally, in 'precal', when you want to find the min or max of a quadratic function, you find the vertex. The vertex is (-b/2a) along with its function value. If the parabola opens upward the vertex is the min; if the parabola opens downward the vertex is the max. But this one opens upward and you asked about the max, so the max would be one of the endpoints.
f(0) = 5
f(4) = -3

Answer 2

find second derivative of the function, set to 0, find x values, plug in humbers to the first derivative, values should be between the x values found......

briefl graph.... form post to neg, is max, from neg to post, is min.

chack in the internet, this topic is easy