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Let $f(x)=(3\sqrt x - \frac 1{\sqrt x})^2 + 4$

Q: Find the value of x when $f(x)=4$ , I thought $x$ could just be $0$, is this correct?

The next thing is, express $f(x)$ in the form $Ax + B + C/X$, where $A, B$ and $C$ are integers and hence find $f'(x)$. I assume the final part is differentiate but could someone point me in the right direction for the first part please?

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    Setting $x=0$ would lead to a division by zero in $\frac{1}{\sqrt x}$, so that is certainly not a solution. – 2017-01-09
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    For the second part, the _first_ step is to multiply out $(3\sqrt x-\frac1{\sqrt x})^2$, which makes the square roots go away. – 2017-01-09

5 Answers 5

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If $f(x)=4$, then

$$(3x^{1/2}-x^{-1/2})^2=0\implies3x^{1/2}=x^{-1/2}$$

Multiply both sides by $x^{1/2}$ to get

$$3x=1$$

$$x=1/3$$

Checking back, this is the solution.

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    Why can the two within the parentheses be equal to each other? – 2017-01-09
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    @user405455 take the square root of both sides. – 2017-01-09
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Considering that $x>0$ $$f(x)=4 \iff (3\sqrt{x}-\frac{1}{\sqrt{x}})^2+4=4 \iff 3\sqrt{x}-\frac{1}{\sqrt{x}}=0 \iff 3x-1=0$$ so: $$f(x)=4 \iff x=\frac{1}{3}$$

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you have to solve $$4=(3x^{1/2}-x^{-1/2})^2+4$$ subtracting $4$ we have to solve $$3x^{1/2}=x^{-1/2}$$ this is equivalent to $x=\frac{1}{3}$

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    what is wrong with THIS answer? – 2017-01-09
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    i have had a simple typo,sorry – 2017-01-09
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$f(x)=4$ means $\displaystyle 3\sqrt{x}=\frac{1}{\sqrt{x}}$,hence $x=\displaystyle\frac{1}{3}$.

for $\displaystyle f(x)=9x+\frac{1}{x}-2$,so $A,B,C$ are pretty obviously.

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$$ 4=(3\sqrt{x}-\frac {1}{\sqrt {x}})^2 +4$$ $$\Rightarrow (3\sqrt {x} -\frac {1}{\sqrt {x}})^2 = 0$$ $$\Rightarrow 3\sqrt {x} = \frac {1}{\sqrt {x}} $$ $$\Rightarrow 3x =1 \Rightarrow x=\frac {1}{3} $$

There is thus no need to expand $f (x) $ and then solve, we just equate RHS and LHS and then we easily get the answer. Hope it helps.

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    This is basically my answer ten minutes late with slightly less explanation. – 2017-01-09