Think about it:
f(x) = 0
x2 = f(x)
Thus:
x2 = 0
x = 0
There is only one root after extracting root.
F(x)=[x^2]+1
So what is the question?
14x
f(x) = x2 + 5x + 1 The roots of this equation are x = -0.2087 and x = -4.7913 (approx).
that isn't a real question.
-2, 1.74 and 0.46
yup
F(x)=[x^2]+1
14x
So what is the question?
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
The zeros of f(x), a function of the variable x, are those values of x for which f(x) = 0. These are points at which the graph of f(x) crosses (or touches) the x-axis. Many functions will do so several times over the relevant domain and the values (of x) are the distinct zeros.
2,025,000.00000... (as many zeros after the decimal point as you want)
f(x) = x2 + 5x + 1 The roots of this equation are x = -0.2087 and x = -4.7913 (approx).
100 deg F = 37.77... deg F
F = ma Force (in Newtons) equals mass (in kilograms) times acceleration (in meters per second squared) In this case, 450 = 30a, so the accelerating is 15 meters per second squared
0 degrees F equals -17.8C