Find the maximum or minimum value of f(x) = 2x^2 + 3x - 5
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calendarEducator since 2010
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We have f(x) = 2x^2 + 3x - 5
At the point of maximum or minimum f'(x) = 0
f'(x) = 4x + 3
4x + 3 = 0
=> x = -3/4
f''(x) = 4, which is positive.
Therefore we get a minimum value at x = -3/4
f(-3/4) = 2*(-3/4)^2 - 3*(3/4) - 5
=> -49/8
The minimum value of the function is -49/8
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calendarEducator since 2008
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f(x) = 2x^2 + 3x -5
First we notice that the function has a positive coefficient for x^2.
Then the fucntion has a minimum values.
Now we will find the derivatives zero.
==> f'(x) = 4x +3 = 0
==> 4x = -3
==> x = -3/4
==> f(-3/4)= 2(-3/4)^2 + 3(-3/4) -5
= 2* 9/16 - 9/4 - 5
= (18- 36 - 80)/16 = -98/16 = -49/8.
==> The minimum values is f(-3/4) = -49/8.
The extreme point of a function is reached if the value of x is cancelling out the 1st derivative.
We'll differentiate the function:
f'(x) = 4x + 3
We'll put f'(x) = 0:
4x + 3 = 0
4x = -3
x = -3/4
The critical point of the function is x = -3/4
We'll calculate the 2nd derivative to decide if the point is maximum or minimum:
f"(x) = 4>0
Since f"(x)>0, then the extreme point of the function is a minimum point.
We'll substitute x by the value of the critical point:f(-3/4) = 2*9/16 - 9/4 - 5
f(-3/4) = 18/16 - 36/16 - 80/16
f(-3/4) = -98/16
f(-3/4) = -49/8
The coordinate of the minimum point of the function are: (-3/4 ; -49/8).
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