There is only one critical point, is it a max or min? Take the second derivative...
f´´(x) = -4
This is negative everywhere, meaning the critical point is a maximum
f(2) = -2(2)² + 8(2) - 3
f(2) = 5
Maximum.....(2, 5) <===================
In case you're not in calculus yet, you could look at this equation and recognize it is a parabola (one x² term) and downward facing (the first term is negative). You then know it has one maximum and no minimums, that max is the vertex, complete the square to find the vertex.
y = -2x² + 8x - 3
y = -2(x² - 4x) - 3
y = -2(x² - 4x + 4) - 3 - 4(-2)
y = -2(x - 2)² + 5
Vertex = (2, 5) <=========
?
2017-01-12 08:30:55 UTC
this is a quadratic equation in a form y = ax²-bx+c, the place a >0, so, the concavity of parabola is up ( parabola is smiling ), so, if the area is in reals, there is not a minimun, there's a optimum.you could calculate the y vertex straight away with the aid of (-delta)/(4a). hence: - delta = - (8*8 - 4*2*15) = fifty six 4a = 4*2 = 8 So the optimum (Y vertex ) is fifty six / 8 = 7 Or any incorrect way is: you will locate the x that correspond the y optimum(the fee of x for which max/min fee occurs).so which you're able to do -b / 2a. hence: -b/2a = -(-8 /2*2 ) = 2 changing this fee in unique equation (2x²-8x+15), you will locate the Yvertex, this is 7, how we calculated before.
moe
2011-02-22 08:12:15 UTC
Rule of thumb:
If f’(x) = 0 and f”(x) > 0 then f(x) is a relative minimum (concave up)
If f’(x) = 0 and f”(x) < 0 then f(x) is a relative maximum (concave down)
First find the first and the second derivaties of the given function f(x):
f'(x) = -4x+8
f"(x) = -4
F’(x) = -4x+8 = 0 (=> x = 2) and F”(x) = -4 (not>0) hence not minimum
on the other hand
F”(x) = -4 which is less than zero (<0) hence it is maximum at x = 2
substitute the value of x (=20, we get:
f(2) = -2*2^2 + 8(2) -3
= -8+16-3 = 5 = answer
William B
2011-02-22 07:35:32 UTC
It would be at the vertex.
x=-b/2a =-8/2(-2) =2
f(2) =-2(2^2) +8(2)-3=
-8 +16-3=5
the maximum value is 5
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