Question

How do you find the max or minimum of `f(x)=2x^2-3x+2`?

Answer 1

Minimum at `:x=0.75` and the minimum value is `0.875`

Explanation:

`f(x)= 2x^2-3x+2 :. f^'(x)= 4x -3` . Critical point is at

`4x-3=0 :. x=3/4=0.75 :.` critical point at `x=0.75`

Draw a number line ,marking critical point at `0.75`

We will examine value of `f"(x)` at one point left of c.p `x=0` and

one point at right of c.p (x=1.0)#

`f^'(0) = 4x-3=4*0-3 = -3 (<0) :.` slope is decreasing and

`f^'(1) = 4x-3=4*1-3 = 1(>0) :.` slope is increasing .

When slope goes from decreasing to increasing the c.p is minimum

point at `:.x=0.75` and the minimum value is

`f(0.75) = 2x^2-3x+2 =2*0.75^2-3*0.75+2 =0.875`

graph{2x^2-3x+2 [-11.25, 11.25, -5.625, 5.625]} [Ans]