гдз 10 клас математика Істер 2018 вправа 18.8
10 клас ➠ математика ➠ Істер
Вправа 18.8
Умова:
Знайдіть приріст функції f(x) = cosx у точці х0 = π/2, якщо:
1) Δх = π/4; 2) Δх = -π/6.
Відповідь:
f(x) = cosx, х0 = π/2
1) Δх = π/4
Δf = f(х0 + Δх) - f(х0)
f(х0) = f(π/2) = cos π/2 = 0
х0 + Δх = π/2 + π/4 = (2π+π)/4 = 3π/4
f(х0 + Δх) = f(3π/4) = cos3π/4 = cos(π - π/4) = -cosπ/4 = -√2/2
Δf = -√2/2 - 0 = -√2/2;
2) Δх = -π/6
Δf = f(х0 + Δх) - f(х0)
f(х0) = f(π/2) = cos π/2 = 0
х0 + Δх = π/2 - π/6 = (3π-π)/6 = 2π/6 = π/3
f(х0 + Δх) = f(π/3) = cos π/3 = 1/2
Δf = 1/2 - 0 = 1/2.
Умова:
Знайдіть приріст функції f(x) = cosx у точці х0 = π/2, якщо:
1) Δх = π/4; 2) Δх = -π/6.
Відповідь:
f(x) = cosx, х0 = π/2
1) Δх = π/4
Δf = f(х0 + Δх) - f(х0)
f(х0) = f(π/2) = cos π/2 = 0
х0 + Δх = π/2 + π/4 = (2π+π)/4 = 3π/4
f(х0 + Δх) = f(3π/4) = cos3π/4 = cos(π - π/4) = -cosπ/4 = -√2/2
Δf = -√2/2 - 0 = -√2/2;
2) Δх = -π/6
Δf = f(х0 + Δх) - f(х0)
f(х0) = f(π/2) = cos π/2 = 0
х0 + Δх = π/2 - π/6 = (3π-π)/6 = 2π/6 = π/3
f(х0 + Δх) = f(π/3) = cos π/3 = 1/2
Δf = 1/2 - 0 = 1/2.