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Solve Lesson 6 Page 43 SBT Math 10 – Kite>

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Topic

Given the function variation table \(y = f\left( x \right)\) as follows:

a) Find the covariate and inverse range of the function \(y = f\left( x \right)\)

b) Compare \(f\left( { – 2021} \right)\) and \(f\left( { – 1} \right)\); \(f\left( {\sqrt 3 } \right)\) and \(f\left( 2 \right)\)

Solution method – See details

On \(\left( {a;b} \right)\) , observe the arrow direction in the variation table

+ The function graph goes up from left to right, the function \(f\left( x \right)\) covariates on \(\left( {a;b} \right)\)

+ The function graph goes down from left to right, the function \(f\left( x \right)\) is inverse on \(\left( {a;b} \right)\)

Detailed explanation

a) Looking at the variation table we see:

Graph of the function going up (from left to right) on \(\left( {1;3} \right)\)

Graph the function going down (from left to right) over two intervals \(\left( { – \infty ;1} \right)\) and \(\left( {3; + \infty } \right)\)

Therefore: The function is covariant over the interval \(\left( {1;3} \right)\) and inversely variable over the interval \(\left( { – \infty ;1} \right) \cup \left( { 3; + \infty } \right)\).

b)

+ Since the function is inverse on the interval \(\left( { – \infty ;1} \right)\), for \( – 2021 < – 1\) we have \(f\left( { – 2021} \right ) > f\left( { – 1} \right)\)

+ Since the function is covariant on the interval \(\left( {1;3} \right)\), for \(\sqrt 3 < 2\) we have: \(f\left( {\sqrt 3 } \right ) < f\left( 2 \right)\)

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