+4 votes
436 views

The function $f (x) = 2.5 \log_e \left( 2 + \exp \left( x^2 - 4x + 5 \right)\right)$ attains a minimum at $x =$?

1. $0$
2. $1$
3. $2$
4. $3$
5. $4$
asked in Calculus
edited | 436 views

## 3 Answers

+11 votes
Best answer
Digvijay is right, $f(x)$ will be minimum at $x = 2,$
Here is another approach.
Since log and exponent are monotonically increasing functions, the problem of minimizing $f(x)$ can be reduced to just minimizing the quadratic expression

$x^{2} - 4x + 5,$
this  quadratic expression can be written as $(x^{2} - 4x + 4) + 1$ which is equal to $(x - 2)^{2} + 1.$
now since $(x - 2)^{2}$ can not be less than $0$, so $(x - 2)^{2} + 1$ can not be less than $1.$
Also $(x - 2)^{2} + 1$ will be at its minimum value $(= 1)$, when $x = 2.$

so value of $f(x)$ will be minimum at $x = 2.$
answered by Boss (14k points)
edited
0
Nice one :)
+3 votes
Just differentiate whole function and put value =0, x comes out to be 2.
1nd derivative +ve here, so minima
answered by Veteran (55.9k points)
edited
+1 vote

In this question it is not necessary to either use differentiation or anything much different.

Just see that how this expression simplifies =>

After cancelling out log & Exponent we get something like =>

2.5ln2 + 2.5 ( x2-4x+5 )

Now 2.5ln2 will be there same in all x values.

Just put in all values one by one, it is not hard to see that , At x = 2 we get minimum value !

Answer c

answered by Boss (43.5k points)
+1

I don't understood how you cancelled log.

After cancelling out log & Exponent we get something like =>

2.5ln2 + 2.5 ( x2-4x+5 )

+7
^
This solution is wrong, He assumed $log(a+b) = log(a)+log(b)$ which is clerly wrong.

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