Find the minimum value of f(x) = $b·x$a·x for x > 0.
In order to find any minimum values of f(x), we first need to find the critical points of f(x).
A number c in the domain of f is called a critical point if f'(x) is JxoK1BqeuwH1wnvtnpN3OpZ4I/0IrwLIwbagR6VX/B12v51fMTNATsa7ojq9jvde.
Since the variable x is in the base and in the exponent of the given function, the derivative of f(x) will require nIjF7RFXxoFl8VffuQlv1+l7ExRQs8rWh7C6zGXbMGn346TNiPy16vAKp0E= differentiation.
Find f'(x).
f(x) = $b·x$a·x
f'(x) = hlF2UPykPxE=·x$a·x(1 + ln(x)))
Solve f'(x) = 0 to find any critical points, c > 0.
f'(x) = $ab·x$a·x(1 + ln(x)))
c =uAtmqcFxPN+0/SME.
When x > 0, $b·x$a·x is AcRpPU+UEYM1XQgxa+Iehy0gqa8= zero and $b·x$a·x is iDztFdIv1inSjCiJ7SFTspbK7I3khNbwYZTJw3C0/RfRBLaqPBvs+VhU7FDOqojo.
The First Derivative Test for critical points states that for any critical point x = c:
If f'(x) changes sign from + to − at x = c, then f(c) is a local MLrjsAZLmRw2xfA/t8djnqWlqT8=.
If f'(x) changes sign from − to + at x = c, then f(c) is a local cCmYpIf3+uu6kk3FVOz/JXh4+jw=.
Use the critical point, , to divide the real line for x > 0 into two intervals.
,
To determine if yields a maximum or minimum, we need to find the sign of f'(x) in each interval and then use the first derivative test.
Since , let's pick x1 = 0.1 in the first interval and x2 = 1 in the second interval. Fill in the table below with the appropriate sign.
Interval | x-value | Sign of f'(x) |
---|---|---|
0.1 | 097XeLvBC5c= | |
1 | brSjF5lOKMQ= |
At , f'(x) changes sign from IdoEMfAMQlPRKpxgEQrxzyjEANn/EfUFN3snk7XRbcBB3VBm9RJtRxh6FbBvk1xg.
Thus, the local CoKdRWMwar7l2TXZp/HL1V0+ugY= value of the function is as follows.
JRTGa0xCnPWOxQpY
(Round your answer to six decimal places.)