When you look at the parabola if it opens downwards then the parabola has a maximum value (because it is the highest point on the graph) if it opens upward then the parabola has a minimum value (because it's the lowest possible point on the graph)
I only know what mean is, so mean is the same thing to the average. * * * * * Range is the difference between the maximum value and the minimum value. Range = Maximum - Minimum.
You set the derivative equal to zero and solve the equation. For example y = x^2 + 5x +7 is the equation of a parabola. dy/dx = 2x +5. Setting 2x +5 = 0 then x = -5/2. When x = -5/2 y = 3/4. This is the minimum. We know it's a minimum and not a maximum because when x is large y is large.
So that's y=-3(x-5)^2 + 4? Okay, so I don't know what you mean by optimum value, but that's going to be y= -3x^2 + 10x - 71, which is an upside down parabola. The maximum of said parabola is going to be where the derivative equals 0, so -6x + 10 = 0, so x= 5/3. Then the corresponding y value is 25/3 - 71... I'll let you do that last bit of math. The maximum of the parabola is then at the point (5/3, 25/3 - 71).
As we know that electric flux is the total number of electric lines of forces passing through a surface. Maximum Flux: Electric flux through a surface will be maximum when electric lines of forces are perpendicular to the surface. Minimum flux: Electric flux through a surface will be minimum or zero when electric lines of forces are parallel to the surface.
Usually at the minimum or maximum of a function, one of the following conditions arises:The derivative is zero.The derivative is undefined.The point is at the end-points of the domain that is being considered (or of the naturally-defined domain, for example, zero for the square root).This will give you "candidate points"; to find out whether each of these candidate points actually is a maximum or a minimum, additional analysis is required. For example, if the second derivative is positive, you have a minimum, if the second derivative is negative, you have a maximum - but if it is zero, it may be a maximum, a minimum, or neither.
In Calculus, to find the maximum and minimum value, you first take the derivative of the function then find the zeroes or the roots of it. Once you have the roots, you can just simply plug in the x value to the original function where y is the maximum or minimum value. To know if its a maximum or minimum value, simply do your number line to check. the x and y are now your max/min points/ coordinates.
I only know what mean is, so mean is the same thing to the average. * * * * * Range is the difference between the maximum value and the minimum value. Range = Maximum - Minimum.
You set the derivative equal to zero and solve the equation. For example y = x^2 + 5x +7 is the equation of a parabola. dy/dx = 2x +5. Setting 2x +5 = 0 then x = -5/2. When x = -5/2 y = 3/4. This is the minimum. We know it's a minimum and not a maximum because when x is large y is large.
minimum is 7 maximum is I don't know
When the quadratic is written in the form: y = ax2 + bx + c then if a > 0 y has a minimum if a < 0 y has a maximum and if a = 0 y is not a quadratic but y = bx + c, and it is linear. The maximum or minimum is at x = -b/(2a)
the minimum is a day and maximum is eight years.i know im right so dont change or question my answer!!!!
i dont quite know at the moment sorry
So that's y=-3(x-5)^2 + 4? Okay, so I don't know what you mean by optimum value, but that's going to be y= -3x^2 + 10x - 71, which is an upside down parabola. The maximum of said parabola is going to be where the derivative equals 0, so -6x + 10 = 0, so x= 5/3. Then the corresponding y value is 25/3 - 71... I'll let you do that last bit of math. The maximum of the parabola is then at the point (5/3, 25/3 - 71).
Solve for when the first derivative is equal to zero. If you don't know how to take a derivative, then put the equation into the form y = Ax2 + Bx + C. The derivative of this will be 2Ax + B, so at x = -B / (2*A), and y = -B2/(4*A) + C
There are two tests involved in checking for maximum and minimum because, if you only checked for a value of zero for the first derivative, you would only know that the equation has zero slope at that point. You also need to check the second derivative to see if that point is a maximum, a minimum, or an inflection point.To be fully correct, you also need to understand the equation itself, because there may be more than one maxima or minima, and/or there may be a discontinuity. This is all part of the process of finding a maximum or minimum.
As we know that electric flux is the total number of electric lines of forces passing through a surface. Maximum Flux: Electric flux through a surface will be maximum when electric lines of forces are perpendicular to the surface. Minimum flux: Electric flux through a surface will be minimum or zero when electric lines of forces are parallel to the surface.
Usually at the minimum or maximum of a function, one of the following conditions arises:The derivative is zero.The derivative is undefined.The point is at the end-points of the domain that is being considered (or of the naturally-defined domain, for example, zero for the square root).This will give you "candidate points"; to find out whether each of these candidate points actually is a maximum or a minimum, additional analysis is required. For example, if the second derivative is positive, you have a minimum, if the second derivative is negative, you have a maximum - but if it is zero, it may be a maximum, a minimum, or neither.