find the derivative of x^3-6x^2+15x+8 and show by completing the square that function has no turning point.
I can find derivative but not sure of next step
To find the derivative of the function f(x) = x^3-6x^2+15x+8, we can use the power rule for differentiation.
The power rule states that if you have a term of the form ax^n, the derivative is given by nx^(n-1).
Applying the power rule to each term in the function f(x), we have:
f'(x) = d/dx (x^3) - d/dx (6x^2) + d/dx (15x) + d/dx (8)
= 3x^2 - 12x + 15
So, the derivative of the function f(x) is f'(x) = 3x^2 - 12x + 15.
Now, let's proceed to show that the function has no turning point by completing the square.
To complete the square, we need to rewrite the quadratic term as a perfect square trinomial.
The quadratic term in our function f'(x) = 3x^2 - 12x + 15 is 3x^2 - 12x.
To complete the square, we can take half of the coefficient of the x-term (-12x), square it, and add it to the expression.
Half of -12x is -6x. Squaring -6x gives us (-6x)^2 = 36x^2.
Adding 36x^2 to 3x^2 - 12x, we have:
3x^2 - 12x + 36x^2 = 39x^2 - 12x
Now, we can rewrite our expression as:
f'(x) = (39x^2 - 12x) + 15 - 36x^2
= 39x^2 - 12x + 15 - 36x^2
= 3x^2 - 12x + 15
Notice that this is the same expression as f'(x) without the square term.
Since the coefficient of the squared term is positive (3 > 0), and the coefficient of the linear term is non-zero (-12 ≠ 0), the quadratic expression 3x^2 - 12x + 15 has no real roots.
Therefore, the function f(x) = x^3-6x^2+15x+8 has no turning points because its derivative f'(x) = 3x^2 - 12x + 15 does not equal zero.