SOLUTION: Let f(x) = 2x^2 − 9x. Find all extreme values (if any) of f on the interval [0, 9]. Determine at which numbers in the interval these values occur. NOTE: If there is no max

Algebra ->  Average -> SOLUTION: Let f(x) = 2x^2 − 9x. Find all extreme values (if any) of f on the interval [0, 9]. Determine at which numbers in the interval these values occur. NOTE: If there is no max      Log On


   



Question 995804: Let f(x) = 2x^2 − 9x.
Find all extreme values (if any) of f on the interval [0, 9]. Determine at which numbers in the interval these values occur. NOTE: If there is no maximum value (or minimum value), then enter DNE in the corresponding blanks.
Maximum Value:
Minimum Value:

Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Definition: Maximum and Minimum of a function
1. A function f has a maximum at x=a if f%28a%29%3E=f%28x%29 for all x in the domain of f.
2. A function f has a minimum at x=a if f%28a%29%3C=f%28x%29 for all x in the domain of f.
The values of the function for these x-values are called extreme values or extrema.

f%28x%29+=+2x%5E2+-9x here you have a parabola
the least or greatest value of the parabola could be found at the vertex of the parabola (on the axis of symmetry x=-b%2F2a)
f%28x%29+=+2x%5E2+-9x=> a=2 b=+-9=>minimum is at x=-b%2F2a=>x=-%28-9%29%2F%282%2A2%29=>x=9%2F4=>x=2.25
or, find it using derivative:
f'%28x%29+=+4x+-9
0=+4x+-9=>4x=9=>x=9%2F4




Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Let f(x) = 2x^2 − 9x
Then f'(x) = 4x-9
And f"(x) = 4
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Find all extreme values (if any) of f on the interval [0, 9]. Determine at which numbers in the interval these values occur. NOTE: If there is no maximum value (or minimum value), then enter DNE in the corresponding blanks.
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Solve 4x-9 = 0
x = 9/4
Since f"(9/4) = 4 > 0,
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Maximum Value: DNE
Minimum Value: f(9/4) occurs at x = 9/4
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Cheers,
Stan H.
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