Question

Math question

Answer 1

differentiate and equate to zero to find the turning points

Answer 2

furthermore the first derivative of your function is:
\[f'\left( x \right) = 9{x^2} - 16\]

Answer 3

円の円周は直径として直接変化します。円は60フィートの直径と188.4フィートの円周を持っています。

15フィートの直径とする円の円周は何ですか？
47.1フィート
7.5フィート
57.1フィート

Answer 4

your function is an increasing function if the subsequent condition holds:
\[9{x^2} - 16 > 0\]

Answer 5

please solve that inequality

Answer 6

9x^2-16>0
9x^2>0+16
9x^2>16
x>4/3

Answer 7

not this way...

Answer 8

find two solution.. square root is written with positive and negative sign...

Answer 9

what's up with those subscripts being x?

Answer 10

we can rewrite that inequality as below:
\[\left( {3x - 4} \right)\left( {3x + 4} \right) > 0\]
so the starting inequality is equivalent to the subsequent two systems of inequalities:
\[\left\{ \begin{gathered}
3x - 4 > 0 \hfill \\
3x + 4 > 0 \hfill \\
\end{gathered} \right.,\quad \left\{ {\begin{array}{*{20}{c}}
{3x - 4 < 0} \\
{3x + 4 < 0}
\end{array}} \right.\]
please solve them

Answer 11

you could have also have found the turning points and their nature and drawn a rough graph
which might have looked like