Consider the following cubic curve:
f(x)=−x3 −x2 +16x+16
1. Calculate f′(x).
2. Calculate f′′(x).
3. Find the x values such that f(x) = 0.
Assuming you meant:
f(x)=?x^3 ?x^2 +16x+16
f ' (x) = -3x^2 - 2x + 16
f ''(x) = -6x - 2
If you are studying Calculus or Pre-Calculus, you MUST know how to do this.
if f(x) = 0
?x^3 ?x^2 +16x+16 = 0
x^3 + x^2 - 16x - 16 = 0
by grouping,
x^2(x+1) - 16(x+1) = 0
(x+1)(x^2 - 16) = 0
(x+1)(x-4)(x+4) = 0
x = -1, 4, -4
verification:
http://www.wolframalpha.com/input/?i=plot+f(x)%3D%E2%88%92x%5E3+%E2%88%92x%5E2+%2B16x%2B16
Why did you repost this question ??
Just noticed that Steve answered it for you several days ago
http://www.jiskha.com/display.cgi?id=1483135842
To calculate f′(x), we need to find the derivative of the given cubic function f(x)=−x^3−x^2+16x+16.
1. Calculate f′(x):
To find the derivative, we differentiate each term of the function with respect to x.
f′(x) = d/dx[-x^3] + d/dx[-x^2] + d/dx[16x] + d/dx[16]
Differentiating each term:
f′(x) = -3x^2 - 2x + 16
So, f′(x) = -3x^2 - 2x + 16.
2. Calculate f′′(x):
To find the second derivative or f′′(x), we differentiate f′(x).
f′′(x) = d/dx(-3x^2 - 2x + 16)
Differentiating each term:
f′′(x) = -6x - 2
So, f′′(x) = -6x - 2.
3. Find the x values such that f(x) = 0:
To find the x values that make f(x) equal to zero, we need to solve the equation -x^3 - x^2 + 16x + 16 = 0.
One way to find the x values is by using numerical methods like the Newton-Raphson method, the bisection method, or the secant method. These methods iterate to approximate the root of an equation.
Alternatively, we can use a graphing calculator or plot the function to visually determine the x values at the points where the graph intersects the x-axis.
By using either of these methods, we can find the x values that satisfy f(x) = 0 for the given cubic function.