If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$

If $$x\geq 0,$$ what is the smallest value of the function $$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$ I tried doing it by completing the square in numerator and making it of the form $$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$ and then, I put the value of $$x= 0$$ and the answer is coming out to be $$13/6.$$

But the actual answer is $$2.$$

Am I missing something ?

• Take the derivative, set it to zero, find $x$. Use the second derivative to see if it is a local maximum or minimum. Mar 11 '21 at 4:21
• @DavidG.Stork Thank you so much! If you don't mind, can you also tell what's wrong with the way i am approaching it? Mar 11 '21 at 4:26
• You are solving for where the function is $0$. Very different problem. Mar 11 '21 at 4:27
• Got it. Thank you so much! Mar 11 '21 at 4:28
• No derivatives needed. $(a-b)^2 = a^2+b^2-2ab \geq 0$ is enough to find the minimum. Mar 11 '21 at 4:33

After completing the squares you can use the inequality between the arithmetic and geometric mean (AM-GM) in the form $$a^2+b^2 \geq 2ab$$:

Hence,

$$\frac{(2(x+1))^2+3^2}{6(x+1)}\stackrel{AM-GM}{\geq}\frac{2\cdot 2(x+1)\cdot 3}{6(x+1)}=2$$

Equality is reached for $$2(x+1)=3 \Leftrightarrow x = \frac 12$$.

• Oh, that made it much simpler...Thank You Mar 11 '21 at 4:34
• @AhmedFaraz You are welcome and 3fwan :-) Mar 11 '21 at 4:38

The derivative, $$\frac{2}{3}-\frac{3}{2 (x+1)^2}$$ is zero when $$x=1/2$$: Let $$x+1=y^2$$ as $$x\ge0, y^2\ge1$$

$$\dfrac{4x^2+8x+13}{1+x}=\dfrac{4(y^2-1)^2+8(y^2-1)+13}{y^2}=4y^2+\dfrac9{y^2}=\left(2y-\dfrac3y\right)^2+2\cdot2y\cdot\dfrac3y\ge12$$ the equality occurs if $$2y-\dfrac3y=0\iff y^2=\dfrac32\iff x=?$$