Trying find the critical numbers for X^3+X^2-8x
But i cant remember how to simplfy the derivitive of 3X^2+2X-8 can someone help me?
Of course! I can help you simplify the derivative of the function f(x) = x^3 + x^2 - 8x.
To simplify the derivative, we need to find the derivative of each term separately using the power rule.
The power rule states that for any term of the form cx^n, the derivative is given by: d/dx (cx^n) = cnx^(n-1).
Let's find the derivative term by term:
1. Derivative of x^3:
Applying the power rule, we get d/dx (x^3) = 3x^(3-1) = 3x^2.
2. Derivative of x^2:
Applying the power rule, we get d/dx (x^2) = 2x^(2-1) = 2x.
3. Derivative of -8x:
The constant -8 does not involve x, so its derivative is 0.
The derivative of x is 1.
Therefore, the derivative of -8x is -8(1) = -8.
Now, let's combine the derivatives of each term:
d/dx (x^3 + x^2 - 8x)
= d/dx (x^3) + d/dx (x^2) + d/dx (-8x)
= 3x^2 + 2x - 8.
So, the derivative of f(x) = x^3 + x^2 - 8x simplifies to 3x^2 + 2x - 8.
Now you can use this simplified derivative to find the critical numbers for your function. Critical numbers occur where the derivative is equal to zero or undefined. Set the simplified derivative equal to zero and solve for x:
3x^2 + 2x - 8 = 0.
You can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
Once you find the values of x from solving the equation, those will be the critical numbers for your function.