Trivial Properties of a Quadratic

Calculus Level 1

If f ( x ) = x 2 f(x) = x^{2} , then

f − 1 ( x ) = x f^{-1}(x) = x f ( f ( x ) ) = x f(f(x))= \sqrt{x} f ( 324 ) = 18 f(324) = 18 f ′ ( x ) = 2 x f'(x)=2x

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Kevin Mo by Kevin Mo , 19, USA

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3 solutions

Discussions for this problem are now closed

Kevin Mo
Sep 28, 2014

Using the power rule , the derivative of x 2 x^{2} is ( 2 ) x 2 − 1 ⇒ 2 x 1 ⇒ 2 x (2)x^{2-1} \Rightarrow 2x^1 \Rightarrow \boxed{2x} . Ans. \textbf{Ans.}

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Paola Ramírez
Jan 21, 2015

A simpe way to solve is using limits lim ⁡ h → 0 f ( x + h ) − f ( x ) h = ( x + h ) 2 − x 2 h = x 2 + 2 x h + h 2 − x 2 h = 2 x h + h 2 h = 2 x \displaystyle\lim_{h \rightarrow 0} \frac {f(x+h)-f(x)}{h}=\frac{(x+h)^2-x^2}{h}=\frac{x^2+2xh+h^2-x^2}{h}=\frac{2xh+h^2}{h}=\boxed{2x}

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And if you wanted to be even more precise you could use the delta-epsilon proof with limits, that is if you wanted to be more rigorous.

Seth Lovelace - 6 years, 4 months ago

This is actually the definition of a derivative.

Roman Frago - 6 years, 4 months ago
Lu Chee Ket
Feb 6, 2015

ln ⁡ ( f ( x ) ) = 2 ln ⁡ ( x ) 1 f ( x ) ⋅ f ′ ( x ) = 2 x f ′ ( x ) = 2 x ⋅ x 2 = 2 x \begin{aligned} \ln (f(x)) &=& 2 \ln(x) \\ \frac 1 {f(x)} \cdot f'(x) &=& \frac 2 x \\ f'(x) &=& \frac 2 x \cdot x^2 = 2x \\ \end{aligned}

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