Tentukan

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1.turunan pertama dan kedua
2.titik stasioner
3. nilai maksimum dan minimum dari;
f(x)=4cos (4x-pi/2)

Tentukan

Jawaban:

1.

f(x) = 4 cos(4x - frac{pi}{2} ) \ f'(x) = 4. - sin(4x - frac{pi}{2} ) \ = - 4 sin(4x - frac{pi}{2} ) \ f''(x) = 4. - 4 cos(4x - frac{pi}{2} ) \ = - 16 cos(4x - frac{pi}{2} )

2.

Titik stasioner, maka f'(x) = 0

f'(x) = - 4 sin(4x - frac{pi}{2} ) \ 0 = - 4 sin(4x - frac{pi}{2} ) \ - - - - - - - - div - 4 \ 0 = sin(4x - frac{pi}{2} ) \ sin(0) = sin(4x - frac{pi}{2} ) \ 4x - frac{pi}{2} = 0° + kpi \ 4x = frac{pi}{2} + 0° + kpi \ 4x = frac{pi}{2} + kpi \ - - - - - - - div 4 \ x = frac{1}{8} pi + frac{1}{4} kpi

maka nilai x yang memenuhi persamaan diatas sekaligus menjadi titik stasionernya

x = frac{1}{8} pi.......(k = 0) \ x = frac{3}{8} pi.......(k = 1) \x = frac{5}{8} pi.......(k = 2) \ x = frac{7}{8} pi.......(k = 3) \ x = frac{9}{8} pi.......(k = 4) \ x = frac{11}{8} pi.......(k = 5) \ x = frac{13}{8} pi.......(k = 6) \ x = frac{15}{8} pi.......(k = 6)

3.

Nilai minimum

x = frac{1}{8} pi \ f(x) = 4 cos(4x - frac{pi}{2} ) \ 0 < 4 cos( (4.frac{1}{8}. pi) - frac{pi}{2} ) \ 0 < 4 cos( 0 ) \ 0 < 4..( + )

x = frac{3}{8} pi \ 0 < 4 cos(4. frac{3}{8}pi - frac{pi}{2} ) \ 0 < 4 cos(pi) \ 0 < - 1...( - )

Semoga membantu

Gambar Jawaban