Determine relative maxima or minima?

Question

Determine any relative maxima or minima of the function and the intervals on which the function is increasing or decreasing.

f(x) = x^2 -2x +3

Answer 1

relative maxima and minima are really just the results of the first derivative test so...

first use power rule to find derivitive...
2x-2=dy/dx

then solve for the crit pt(s)

x=1

then test values on either side of the crit pt (I chose 0 and 2)

0-2= a negative number
4x-2=a posivite number

theofore, according to the first derivative test when x=1 b/c f(x) changes from a + to - number f(1) is a minimum value

Answer 2

Given f(x) = x^2 - 2x + 3
Then, f '(x) = 2x - 2; f'(x) is 0 at x = 1, so that is the relative maxima / minima.
One way to find out which is to determine f"(1), which is simply 2. Since f"(1) is positive, the point is a minimum
Since f(x) is a parabola, f(x) decreases for all x < 1 and increases for all x > 1.

Answer 3

Find f'(x), then set f'(x) = 0 to determine the local extrema.

Use f''(x) (second derivative) to find the intervals where the function is increasing (f''(x) > 0) or decreasing (f''(x) < 0).

Answer 4

one million) to locate relative optimal and minimum, use the roots of the 1st spinoff... f´(x) = 4 -2x ... there is in basic terms a root x = 2. learn the sign of f´(x) ... + + + + + 2 - - - - - - Then at x =2 the function has a relative optimal... the ingredient is (2,4). to locate the inflection ingredient locate the muse of the 2d spinoff... f´´(x) = -2 .... the 2d spinoff has no longer a root, then there's no ingredient of inflection... the function is concave down. 2) to locate x-intercept, in basic terms placed y = 0 ... then 0 = 4x -x^2 ... it rather is (0,0) and (2,0) to locate the y-intercept, in basic terms placed x = 0 ... then x=0, y=0... it rather is (0,0). ok!