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Derivative of One Function w.r.t Another

20. Let f(x)=...

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20. Let f(x)=1/x and g(x)=1/x then

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f (x) = [\frac{x}{[x]}], g (x) = [x [\frac{1}{x}]]

h (x) = [\frac{[x]}{x}],

lim x \to 2^{-} f (x) + lim x \to {\frac{1}{2}}^{+} g (x) + lim x \to 2^{+} h (x) =

f (x) = | x - 2 | + | x - 3 | + | x - 4 |

g (x) = f (x + 1)

Q. Assertion :Let

f (x) = (x + 1)^{2} - 1 \forall x \geq - 1

g (x) = - 1 + \sqrt{x + 1}

then number of solutions of the equation

g (x) = f (x)

f (x)

g (x)

are inverse of each other.

x^{2} + \frac{1}{x^{2}}

\frac{1}{x}, x \in R - {- 1, 0, 1} . I F h (x) = \frac{f (x)}{g (x)}, t h e n t h e l o c a l m i n i m u m v a l u e o f h (x) i s :

Q. Assertion :Let the function

f (x) = x^{2} - x + 1 \forall x \geq \frac{1}{2}

g (x) = \frac{1}{2} + \sqrt{x - \frac{3}{4}}

f (x) = g (x)

has two solutions. Reason:

f (x)

g (x)

are inverse of each other.

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