dik fungsi f(x) =-2 sin (2x - 30°) untuk 0° ≤ x ≤180° . grafik fungsi tersebut cekung ke atas pada interval..

<p>f(x) = -2sin (2x - 30°)&nbsp;</p><p>f'(x) = -4 cos (2x - 30°)</p><p>f''(x) = 8 sin (2x - 30°)</p><p>&nbsp;</p><p>suatu fungsi cekung ke atas jika f''(x) &gt; 0.</p><p>&nbsp;</p><p>tentukan nilai x dengan f''(x) = 0</p><p>8 sin (2x - 30°)= 0&nbsp;</p><p>sin(2x - 30°) = 0</p><p>sin (2x - 30°) = sin 0°</p><p>(i) 2x - 30° = 0° + 360°.k</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2x = 30° + 360°.k</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; x = 15° + &nbsp;180°.k</p><p>&nbsp;&nbsp;<strong> &nbsp; &nbsp; k = 0 -&gt; x = 15°</strong></p><p>(ii) &nbsp;2x - 30° = 180° + 360°.k</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2x = 210° + 360°.k</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;x = 105° + &nbsp;180°.k</p><p>&nbsp; &nbsp; &nbsp; &nbsp; k =0 -&gt; x = 105°</p><p><strong>diperoleh x = {15°, 105°}.</strong></p><p>&nbsp;</p><p>Uji titik</p><p>f''(x) = 8 sin (2x - 30°)</p><p>f''(90°) = 8 sin (2(90°)- 30°)</p><p>f''(90°) = 8 sin (180°- 30°)</p><p>f''(90°) = 8 sin (150°)</p><p>f''(90°) = 8 x 1/2</p><p>f''(90°) = 4 (tandanya positif)</p><p>&nbsp;</p><p>sehingga pada interval,</p><p>0° ≤ x &lt; 15° : cekung ke bawah (tandanya -)</p><p>15° &lt; x &lt; 105° : cekung ke atas (tandanya +)</p><p>105° &lt; x ≤ 180° : cekung ke atas (tandanya -)</p><p>&nbsp;</p><p>Maka, grafik fungsi tersebut cekung ke atas pada interval 15° &lt; x &lt; 105°.</p>