max/min values?

Question

for each function state
a) the max/min value
b) The value of x for which max/min value occurs

y=2x²-8x+15

Answer 1

i dont know what level you are? if you are in calculus you can use derivative to do it
if you are on precal or below level you can use vertex to do .


this graph opens upward because a = 2> 0 so we have minimum value


vertex : x = -b/2a , y = f(-b/2a)

y=2x²-8x+15

a = 2, b = -8, c = 15
x = -b/2a = --8/2*2 = 2

y = 2(2)^2-8(2)+ 15 = 7


by calculus way: take derivative with respect to x

y' = 4x -8

set y' = 0 tofigure critical point

4x-8= 0

4x= 8
x = 2

take second derivative with respect to x
y" = 4 > 0 therefore this graph will be concaving up , we dont have maximum point but we have minimum point at x = 2

put x = 2 to orginal question to get y

y = 2(2)^2-8(2)+15 = 7

Answer 2

It is a quadratic equation in a form y = ax²-bx+c, where a >0, so, the concavity of parabola is up ( parabola is smiling ), so, if the domain is in reals, there is not a minimun, there is a maximum.You can calculate the y vertex directly by (-delta)/(4a).
in this case:
- delta = - (8*8 - 4*2*15) = 56
4a = 4*2 = 8
So the maximum (Y vertex ) is 56 / 8 = 7

Or the other way is:
you can find the x that correspond the y maximum(The value of x for which max/min value occurs).So you can do -b / 2a. In this case:
-b/2a = -(-8 /2*2 ) = 2
replacing this value in original equation (2x²-8x+15), you will find the Yvertex, that is 7, how we calculated before.

Answer 3

One way to find critical values (maximum, minimum, inflection) is by differentiating.
dy = 4x*dx -8*dx
dy/dx = 4x - 8

dy/dx (the rate at which y changes, compared to changes in x) is the slope of the curve. The slope changes at various points (because the "curve" is not a straight line -- in this case, it is a parabola).

At the exact point where the curve reaches a maximum, the slope is zero. It must be. If the curve had been going up, then if the slope is still positive, the curve continues to go up (we are not yeat at the maximum). If it is now negative, we have past the maximum (we are already on the way down).

So, set dy/dx equal to zero and solve for x:

0 = 4x - 8
8 = 4x
8/4 = x = +2

The critical point is at x = +2

To find the value of the equation at the critical point, replace x with 2 in the original equation:

y = 2x²-8x+15 = 8 - 16 + 15 = +7

The critical point is at (2, 7)

We have not yet established if it is a max, a min or something else (an inflection point).

The "easy" way is to take the second differential:

d(dy/dx) = 4*dx
d²y/dx² = +4
This is the rate of change of the slope (the slope is the rate of change of the equation). This means that the slope keeps increasing.
Therefore, just before the slope (dy/dx) reached a value of 0, it must have been negative (the original curve was going down); just after the slope reached zero, it became positive (the original curve started back up). Therefore our critical point was a minimum.

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Critical points:

You are looking for critical points in a curve.

Take the first differential and set it equal to 0.
All answers (values of x) are critical point.
Let us say that one of the values is "k" (x = k)

Take the second differential and evaluate it at the point x=k. If the result is positive, the critical point is a minimum.
If the result is negative, the critical point is a maximum.
If the result is zero, the critical point is an inflection point.

Answer 4

There is no maximum for this function since as x--> infinity y --> infinity.

The minimum occurs when the derivative is 0. The derivative of 2x²-8x+15 is 4x-8 which is 0 when x=2. The value of y when x=2 is 7.

If you don't know derivatives just graph the function - you'll see that the function has a minimum at (2, 7)

Answer 5

dy/dx= 4x-8
4x-8=0 at x=2 the value for which maximum occurs and the only min/max

the max value is 2(2)²-8(2)+15= 7

Answer 6

take the derivative

y'=4x-8
y''=4

since the second derivative is positive, the function is concave up.

0=4x-8 x=2

it has min value since concave up at x=2 and the value is 7

Answer 7

You made an error when finding the derivative. It's -(x^2-16) on top. Your critical numbers are plus and minus 4.