Himpunan penyelesaian dari pertidaksamaan eksponen...

Himpunan penyelesaian dari pertidaksamaan eksponen $open parentheses 1 over 8 close parentheses to the power of x plus 3 end exponent greater or equal than open parentheses 1 half close parentheses to the power of x squared minus 1 end exponent$ adalah ....

$open curly brackets x vertical line x greater or equal than 7 1 half close curly brackets$
$open curly brackets x vertical line x greater or equal than 3 over 10 close curly brackets$
$open curly brackets x vertical line x greater or equal than negative 13 over 10 close curly brackets$
$open curly brackets x vertical line 2 less than x less than 4 close curly brackets$
$open curly brackets x vertical line minus 2 less or equal than x less or equal than 5 close curly brackets$

Jawaban:

$open parentheses 1 over 8 close parentheses to the power of x plus 3 end exponent greater or equal than open parentheses 1 half close parentheses to the power of x squared minus 1 end exponent open parentheses 1 half close parentheses to the power of 3 x plus 9 end exponent greater or equal than open parentheses 1 half close parentheses to the power of x squared minus 1 end exponent 3 x plus 9 less or equal than x squared minus 1 x squared minus 3 x minus 10 greater or equal than 0 left parenthesis x plus 2 right parenthesis left parenthesis x minus 5 right parenthesis equals 0 x equals negative 2 space logical or space x equals 5$

Himpunan penyelesaian pertidaksamaan tersebut dapat diketahui dengan melakukan substitusi bilangan-bilangan

di sekitar -2 dan 5 ke $open parentheses 1 over 8 close parentheses to the power of x plus 3 end exponent minus open parentheses 1 half close parentheses to the power of x squared minus 1 end exponent greater or equal than 0$

Perhatikan gambar garis bilangan berikut!

Dengan demikian, himpunan penyelesaian dari pertidaksamaan tersebut adalah..

HP= $open curly brackets x vertical line minus 2 less or equal than x less or equal than 5 close curly brackets$

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