Question

f(x)=1+3x-x^2

find the min and max values of the function

Answer 1

this is a parabola that faces down, so it has no minimum
maximum is the second coordinate of the vertex
first coordinate of the vertex is always \(-\frac{b}{2a}\)
in your case it is \(-\frac{3}{2\times (-1)}=\frac{3}{2}\)
replace \(x\) by \(\frac{3}{2}\) to get the maximum

Answer 2

isnt that 3/-2??

Answer 3

Do you want a calculus based or a non-calculus based solution?

Answer 4

non calc based

Answer 5

Alright!

First rearrange the original polynomial to the form

f(x) = ax² + bx + c

Hence

f(x) = -x² + 3x + 1

Now complete the square

f(x) = (-x² + 3x) + 1 [step 1: separate the variable terms from the constant]

f(x) = -(x² - 3x) + 1 [step 2: factor the coefficient of x² from the parentheses]

f(x) = -(x² - 3x + 9/4 - 9/4) + 1 {step 3:add and subtract the square of half the coefficient of x inside the parentheses]

f(x) = -(x² - 3x + 9/4) + (-9/4)(-1) + 1 [step 4: remove the -9/4 from the parentheses and multiply by the original coefficient of x² that was factored in step 2]

f(x) = -(x - 3/2)² + 13/4 [step 5: a² - 2ab + b² = (a - b)²]

The vertex of this parabola is at (3/2, 13/4) and the parabola's direction of opening is downward.

∴ the maximum value is 13/4 and occurs at x = 3/2.


Note: It is not necessary to perform all 5 steps of completing the square, however when I'm teaching, I like explaining everything properly.

I really hope that helps!

Answer 6

no it is \(\frac{3}{2}\) because there is a minus sign in the formula, and also \(a=-1\)

Answer 7

The maximum value is NOT 3/2. An optimum value is the y-coordinate of the vertex and occurs at the x-coordinate of the vertex. THEREFORE THE MAXIMUM VALUE IS 13/4 WHEN x = 3/2.