## Solve Lesson 6 Page 43 SBT Math 10 – Kite>

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: $$f\left( {\sqrt 3 } \right ) < f\left( 2 \right)$$