John asked in Science & MathematicsMathematics · 4 weeks ago

# Find the maximum of f(x)=2+5x−x^2?

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• Como
Lv 7
4 weeks ago

:-

f ' (x) = 5 - 2x = 0 for turning point

x = 5/2

f (5/2) = 2 + 25/2 - 25/4 = 2 + 25/4 = 33/4

Turning point (5/2 , 33/4)

f ' ' ( x ) = - 2 < 0 thus is a maximum turning point.

• 4 weeks ago

f '(x) = 5 - 2x

0 = 5 - 2x

2x = 5

x = 5/2

f ''(x) = -10 < 0 so we know a maximum occurs at x = 5/2.

f(1/2) = 2 + 5(5/2) - (5/2)^2 = 2 + 25/2 - 25/4 = 8/4 + 50/4 - 25/4 = 33/4

• MyRank
Lv 6
4 weeks ago

f(x) = 2 + 5x - x²

2 + 5x - x² = 0x = -5 ± √25 - 4.2(-1) / 2 . -1= -5 ± √25 + 8 / -2= -5 ± √33 / -2f’(x) = 0 + 5.1 - 2x= 5 - 2xf’(x) = 05 - 2x = 0x = 5/2f(x) > 0 minimum valuef(x) < 0 maximum value.

• 4 weeks ago

The maximum of f(x) = 2 + 5x − x^2 or -x^2 + 5x + 2

Since this is a parabolic equation and the value of #a < 0#

it opens downwards so it has an absolute maxima.

The maximum point is determined by #x_(max) = -b/(2a)#

where b and a are coefficients.

-5/-2 = 2.5

sub x = 2.5 back into equation to get y = 8.25

• 4 weeks ago

without calculus, first write it in Standard Form:

f(x)=−x^2 +5x +2 is the equation in the correct 'form'

From the Quadratic Equation: use -b/2a  ±√(b² -4ac)/2a

x = -b/2a = -(5)/(2*-1) = 5/2

sub x=5/2 back into equation to get y=8.25

With calculus: take 1st derivative = 0, that's where slope =0, for the minimum or maximum.  -2x +5 = 0, then x=5/2

• Vaman
Lv 7
4 weeks ago

f(x)= 2+5x-x^2. To find a maximum or minimum, take fist derivative and equate that to 0. df/dx= 5- 2x. x=5/2, take the second derivative, if it is positive for x, then it is a minimum and of negative then it is a maximum. Here d^2f/dx^2=-5. The point x=5/2 is a maximum

f= 1+25/2-25/4=27/2-25/4= (54-25)/2=29/2. This is the maximum value.

• ted s
Lv 7
4 weeks ago

IF you have studied parabola properties you should know that the vertex is when x = - b / 2a....here x = 5/2...so find f(5/2)...