Question

What are the possible number of positive, negative, and complex zeros of f(x) = 3x4 - 5x3 - x2 - 8x + 4 ?

Answer 1

Hoping you know derivatives!
Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(3 x^4-5 x^3-x^2-8 x+4)
Rewrite the expression: d/dx(f(x)) = d/dx(f(x)):
d/dx(f(x)) = d/dx(4-8 x-x^2-5 x^3+3 x^4)
The derivative of f(x) is f'(x):
f'(x) = d/dx(4-8 x-x^2-5 x^3+3 x^4)
Differentiate the sum term by term and factor out constants:
f'(x) = d/dx(4)-8 d/dx(x)-d/dx(x^2)-5 d/dx(x^3)+3 d/dx(x^4)
The derivative of 4 is zero:
f'(x) = -8 (d/dx(x))-d/dx(x^2)-5 (d/dx(x^3))+3 (d/dx(x^4))+0
Simplify the expression:
f'(x) = -8 (d/dx(x))-d/dx(x^2)-5 (d/dx(x^3))+3 (d/dx(x^4))
The derivative of x is 1:
f'(x) = -(d/dx(x^2))-5 (d/dx(x^3))+3 (d/dx(x^4))-1 8
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
f'(x) = -8-5 (d/dx(x^3))+3 (d/dx(x^4))-2 x
Simplify the expression:
f'(x) = -8-2 x-5 (d/dx(x^3))+3 (d/dx(x^4))
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 3: d/dx(x^3) = 3 x^2:
f'(x) = -8-2 x+3 (d/dx(x^4))-5 3 x^2
Simplify the expression:
f'(x) = -8-2 x-15 x^2+3 (d/dx(x^4))
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 4: d/dx(x^4) = 4 x^3:
f'(x) = -8-2 x-15 x^2+3 4 x^3
Simplify the expression:
f'(x) = -8-2 x-15 x^2+12 x^3
Expand the left hand side:
Answer: |
| f'(x) = -8-2 x-15 x^2+12 x^3