The maximum value of the function
The maximum value of the function
f
(
x
)
=
2
x
3
−
1
5
x
2
+
3
6
x
−
4
8
on the set
A
=
{
x
∣
x
2
+
2
0
≤
9
x
∣
}
is,
The answer is 7.
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2 solutions
Note that you solved or ( x − 5 ) ( x − 4 ) = 0 , not ( x − 5 ) ( x − 4 ) ≤ 0 . In truth, the possible values of x are in the interval x ∈ [ 4 , 5 ] .
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I wrote that solution long ago when I was not good at it.I have edited it now can you check if it is correct?
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Yep, it looks good now ⌣ ¨
x 2 − 9 x + 2 0 ≤ 0
5>X>4
From here, We take the second derivative and plug in both 4 and 5, we see that at Both, the function is concave downwards meaning that the max value of this function occurs at x=5 given this limited range
Thus plugging in x=5, y=7
We can write x 2 − 9 x + 2 0 = ( x − 5 ) ( x − 4 )
Therefore,
( x − 5 ) ( x − 4 ) ≤ 0
Therefore x ∈ [ 4 , 5 ]
f ′ ( x ) = 6 x 2 − 3 0 x + 3 6 > 0 ∀ ( − ∞ , 2 ) ∪ ( 3 , ∞ ) 2 , 3 are roots of f ′ ( x ) .
Therefore we see that f ( x ) is increasing for interval x ∈ [ 4 , 5 ] as [ 4 , 5 ] ∈ ( 3 , ∞ ) .
Therefore f ( 5 ) = 7 is maximum value.