f(x) = x2 + 5x + 1 The roots of this equation are x = -0.2087 and x = -4.7913 (approx).
14x
f = 1
7 + f - 21 = -20 Therefore, 7 + f = 1 f = 1 - 7 f = -6
wth??
f(x) = x2 + 5x + 1 The roots of this equation are x = -0.2087 and x = -4.7913 (approx).
14x
f = 1
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
f^2 + 2f = f (f + 2)
f(x)=x+1 g(f(x))=x f(x)-1=x g(x)=x-1
-2, 1.74 and 0.46
7 + f - 21 = -20 Therefore, 7 + f = 1 f = 1 - 7 f = -6
wth??
yup
∫ f'(x)/(p2 + q2f(x)2) dx = [1/(pq)]arctan(qf(x)/p)
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