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Wherever there is an x substitute -2 and calculate the resulting sum: f(x) = 3x² - x → f(-2) = 3 × (-2)² - (-2) = 3 × 4 + 2 = 12 + 2 = 14
f(x) = √(2x -3) f(x) = (2x - 3)^(1/2) f'(x) = (1/2)[(2x - 3)^(1/2 - 1)](2) f'(x) = (2x - 3)^(-1/2) f'(x) = 1/[(2x - 3)^(1/2)] f'(x) = 1/√(2x -3)
f(x) = 2 cos 3x The amplitude: A = |2| = 2 The maximum value of the function: 2 The minimum value of the function: -2 The range: [-2, 2]
F(x)=x*Sqrt(x^2+1) let f(x)=x let g(x)=Sqrt(x^2+1) F'(x)=f'(x)*g(x)+f(x)*g'(x) f'(x)=1 Use chain rule: g'(x)=(x)*(x^2+1)^(-1/2) F'(x)=Sqrt(x^2+1)+(x^2)/Sqrt(x^2+1)
f(x)=|x| |x| = sqrt(x^2) f(x) = sqrt (x^2) f'(x)=1/2sqrt(x^2) d/dx(x^2) f'(x) = 2x /2sqrt(x^2) f'(x) = x/sqrt(x^2) f'(x)= x /|x| --------------------- Note: d/dx [sqrt(x^2] = d/dx [x^(1/2)] =(1/2)x^(-1/2)=1/2x^(1/2)=1/2sqrt(x)
F 2 cattle is cattle whit f 2
f(x)=x2-x3 f(2) = 4-8 = -4 f'(x) =2x-3x2 f'(2) = 4-12=-8 f''(x) = 2 -6x f''(2)= -10 f'''(x)= -6 f(n)(x) = 0 for all n > 3. f(x) = f(2) + (x-2) f'(2) / 1! + (x-2)2 f''(2) /2! + (x-2)3 f'''(2)/3! + . . . f(x) = -4 -8(x-2) -10(x-2)22/2-6(x-2)3/6 + 0 + 0 + ... f(x) = -4 -8(x-2) -5(x-2)2 - (x-2)3
Are you trying to solve for x? Fx = x2 - 3 x2 - Fx - 3 = 0 x2 - Fx = 3 x2 - Fx + (F/2)2 = 3 + (F/2)2 (x - F/2)2 = 3 + (F/2)2 x - F/2 = ±[ 3 + (F/2)2 ]1/2 x = F/2 ± [ 3 + (F/2)2 ]1/2
if f(x) = 3x + 2 and you're looking for f(2), just plug in 2 for x: f(2) = 3(2) + 2 = 6 + 2 = 8.
To find f(-2), substitute -2 into the function f(x). So, f(-2) = 2(-2)^3 - 2(-2)^2 + 50. Simplifying this expression gives f(-2) = 2(-8) - 2(4) + 50 = -16 - 8 + 50 = 26.
f^2 + 2f = f (f + 2)
f+4(f-2)
f(x) = x2 + 3x - 2 then f'(x) = 2x + 3 and then then f'(2) = 2*2 + 3 = 4+3 = 7
^ means "to the power of" F ^ 2 + 4 = 5 ^ 2 F ^2 +4 = 25 F ^2+4 -4 = 25 -4 F^2 = 21 F = (square root of 21)
F(x) = 3x + 2 f(5) = 3*5 + 2 = 17
f(n) = (n² + n) / 2 f(1) = (1² + 1) / 2 = 1 f(2) = (2² + 2) / 2 = 3 f(3) = (3² + 3) / 2 = 6 f(4) = (4² + 4) / 2 = 10 f(5) = (5² + 5) / 2 = 15 f(6) = (6² + 6) / 2 = 21
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