-2 = - x + x^2 -4 => x^2 - x - 4 + 2 = 0
x^2 - x - 2 = 0
a is the coefficient of x^2 => a = 1
b is the coefficient of x => b = - 1
c is the constant term => c = - 2
quadratic equation: [- b +/- √(b^2 - 4ac) ] / 2a =
= { 1 +/- √[ (-1)^2 - 4(1)(-2)] } / (2(1) = { 1 +/- √ (1 + 8) } / 2 = {1 +/- √9} / 2 =
= { 1 +/- 3} / 2
F(x) = x - 2sinx
To find for what values of x graph has tangents we need to differentiate function f(x).
f'(x) = 1-2cosx
Now, if we want to have horizontal tangents we need to make function f'(x) equal to 0 because at minimum or maximum of function f(x) tangent is horizontal.
1-2cosx = 0
cosx = 1/2
x = pi/3 + 2*n*pi and
x = -pi/3 + 2*n*pi
Explanation: cosx = 1/2 for both 60° and -60°. If we make one circle cosx will be equal to 1/2 for 420 and 300 degrees. Writing degrees in radians (using pi) we can write formulas above where 2*pi represents one full circle and n is an integer which represents how many circles we made.
Answer:
f(g(x)) = 9x^2 + 15x - 6
Step-by-step explanation:
We are using function g(x) = 3x - 1 as the input to function f(x) = x^2 + 7x.
Starting with f(x) = x^2 + 7x, substitute g(x) for x on the left side and likewise substitute x^2 + 7x for each x on the right side. We obtain:
f(g(x)) = (3x - 1)^2 + 7(3x - 1).
If we multiply this out, we get:
f(g(x)) = 9x^2 - 6x + 1 + 21x - 7, or
f(g(x)) = 9x^2 + 15x - 6
