KUIS MATEMATIKA Kode: 01-4-12

GRAFIK FUNGSI

Grafik fungsi bernilai minimum saat f'(x) = 0
[tex]\displaystyle f'(x) = - \frac{1}{2} (-2x) \sqrt{25 - x^{2} } \\ 0 = \frac{x}{ \sqrt{25- x^{2} }} \\ x = 0 f(0) = 7 - \sqrt{25-0^{2}} = 7 - 5 = 2 Titik (0, 2) a^{2} + b^{2} = 0^{2} + 2^{2} = 4 (A) -\ \textgreater \ Jawab [/tex]

titik minimum dpt ditentukan dgn turunan
f'(x) = -1/2 (25-x^2)^-1/2 (-2x) = 0
Diperoleh x = 0.
Substitusikan x = 0 ke pers semula
f(0) = 7 - akar (25-0) = 7 - 5 =2.
Jadi a = 0 dan b = 2 sehingga
a^2 + b^2 = 4.