Hello, yullin!
Find the absolute extreme of the function:
f ( x ) = 3 x 4 x 2 + 1 \displaystyle \,f(x)\:=\:\frac{3x}{\sqrt{4x^2\,+\,1}} f ( x ) = 4 x 2 + 1 3 x on the interval [-1,1].
I am stuck on:
f ′ ( x ) = ( 4 x 2 + 1 ) 1 2 ( 3 ) − ( 3 x ) ( 1 2 ) ( 4 x 2 + 1 ) − 1 2 ( 8 x ) 4 x 2 + 1 \displaystyle \,f'(x)\:=\:\frac{(4x^2+1)^{\frac{1}{2}}(3)\,-\,(3x)(\frac{1}{2})(4x^2+1)^{-\frac{1}{2}}(8x)}{4x^2\,+\,1} f ′ ( x ) = 4 x 2 + 1 ( 4 x 2 + 1 ) 2 1 ( 3 ) − ( 3 x ) ( 2 1 ) ( 4 x 2 + 1 ) − 2 1 ( 8 x )
I bet you
do know what to do next . . . but the algebra stops you, right?
We have: \(\displaystyle \L\,f'(x)\:=\:\frac{3(4x^2+1)^{\frac{1}{2}}\,-\,12x(4x^2+1)^{-\frac{1}{2}}}{4x^2\,+\,1} \;= \;0\)
Multiply top and bottom by
( 4 x 2 + 1 ) 1 2 : \displaystyle (4x^2+1)^{\frac{1}{2}}: ( 4 x 2 + 1 ) 2 1 :
\(\displaystyle \L\;\;\frac{(4x^2+1)^{\frac{1}{2}}}{(4x^2+1)^{\frac{1}{2}}}\,\cdot\,\frac{3(4x^2+1)^{\frac{1}{2}}\,-\,12x(4x^2+1)^{-\frac{1}{2}}}{4x^2\,+\,1}\;=\;0\)
and we get: \(\displaystyle \L\:\frac{3(4x^2+1)\,-\,12x}{(4x^2\,+\,1)^{\frac{3}{2}}} \;=\;\frac{12x^2\,+\,3\,-\,12x}{(4x^2\,+\,1)^{\frac{3}{2}}} \;=\;\frac{3(4x^2\,-\,4x\,+\,1)}{(4x^2\,+\,1)^{\frac{3}{2}}\)
Then: \(\displaystyle \L\,\frac{3(2x\,-\,1)^2}{(4x^2\,+\,1)^{\frac{3}{2}}} \;=\;0\;\;\Rightarrow\;\;x\,=\,\frac{1}{2}\)
When
x = 1 2 , f ( 1 2 ) = 3 2 4 ≈ 1.06 \displaystyle x\,=\,\frac{1}{2},\;f\left(\frac{1}{2}\right)\;=\;\frac{3}{\sqrt{2}}{4}\;\approx\;1.06 x = 2 1 , f ( 2 1 ) = 2 3 4 ≈ 1 . 0 6
Check the endpoints:
\(\displaystyle \L\;\;f(-1)\;=\;\frac{-3}{\sqrt{4\,+\,1}}\;=\;-\frac{3}{\sqrt{5}}\;\approx\;-1.34\;\) . . . absolute minimum
\(\displaystyle \L\;\;f(1)\;=\;\frac{3}\sqrt{4\,+\,1}}\;=\;\frac{3}{\sqrt{5}}\;\approx\;1.34\;\) . . . absolute maximum
Evidently, our critical value does not produce an extreme value.
[I checked it out with the second derivative . . . very messy!
There is a horizontal tangent at
x = 1 2 \displaystyle x\,=\,\frac{1}{2} x = 2 1 , but it's an inflection point.]
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