Number of possible turning points in the following equation
f(x)=x^2(x-3)(x+4)
show work please
Sorry, I only do the non-URGENT ones
look at the factors. THere is a turning point at 3, -4. At zero, f(x) keeps the same curvature.
so does that mean there are 2 turning points
Number of possible turning points in the following equation
f(x)=x^2(x-3)(x+4)
are there 2 turning points or three turning points on the graph
To determine the number of possible turning points in the equation f(x) = x^2(x - 3)(x + 4), you need to analyze the behavior of the function's graph.
Turning points occur where the function changes from increasing to decreasing or vice versa. For a polynomial function, turning points can only occur at the x-values where the derivative of the function equals zero.
To find the derivative of the function, we need to apply the product rule and chain rule:
f'(x) = d/dx [x^2(x - 3)(x + 4)]
= 2x(x - 3)(x + 4) + x^2(1)(x + 4) + x^2(x - 3)(1)
= 2x(x - 3)(x + 4) + x^2(x + 4) + x^2(x - 3)
= 2x(x^2 + 4x - 3x - 12) + x^2(x + 4) + x^2(x - 3)
= 2x(x^2 + x - 12) + x^2(x + 4) + x^2(x - 3)
= 2x^3 + 2x^2 - 24x + x^3 + 4x^2 + x^3 - 3x^2 + x^2
= 4x^3 - 26x^2 - 24x
Now that we have the derivative f'(x), we need to find the x-values where f'(x) = 0:
4x^3 - 26x^2 - 24x = 0
To factor this equation, we can take out the common factor x:
x(4x^2 - 26x - 24) = 0
Now, we have two possibilities:
1) x = 0
2) 4x^2 - 26x - 24 = 0
To solve the quadratic equation, we can factorize or use the quadratic formula. Factoring is not straightforward in this case, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the equation 4x^2 - 26x - 24 = 0, a = 4, b = -26, and c = -24.
Plugging these values into the quadratic formula:
x = (-(-26) ± √((-26)^2 - 4(4)(-24))) / (2(4))
x = (26 ± √(676 + 384)) / 8
x = (26 ± √(1060)) / 8
x = (26 ± √(4 * 265)) / 8
x = (26 ± 2√(265)) / 8
x = (13 ± √(265)) / 4
So, the possible x-values for turning points are x = 0, x = (13 + √(265)) / 4, and x = (13 - √(265)) / 4.
Therefore, there are three possible turning points in the equation f(x) = x^2(x - 3)(x + 4).