Please help! These 2 questions are very similar, and i tried to do completing square method, but i cant get the correct answer! I'm trying to get the dy/dx to be a complete square thingy to show that the value is always greater or equal to 0.
1. Show that the function f(x)= x^3 - 3x^2 + 9x - 5 is increasing with x for all real values of x.
f'(x)=3x^2-6x+9
=3(x^2-2x+3)
=3[x^2 - 2x + (-2/2)^2 + 3 - (-2/2)^2]
= 3[ (x-1)^2 + 2]
=3(x-1)^2 + 6
?!! The correct answer is supposed to be f'(x)=3(x-1)^2 is ≥0 for all real values of x. Did i do the completing square wrongly?? This same thing happened with the second question
2. Show that f(x) = -2x^2 + 3x^2 - 2x + 4 decreases for all real values of x.
The correct answer for this one is f'(x) = -6(x-1/2)^2 - 1/2 <0 for all real values of x
1. You did it correctly.
if the first term (x-1)^2 is positive for all x, then that term +6 is positive for all x.
2. Your f' does not match the degree of f(x). I don't know what you meant.
In both cases, you have made a slight mistake in completing the square. Let's go through the steps for each question to help you correct the errors.
1. Show that the function f(x) = x^3 - 3x^2 + 9x - 5 is increasing with x for all real values of x.
To find the derivative, f'(x), of the function f(x), you correctly calculated:
f'(x) = 3x^2 - 6x + 9
Now, let's complete the square correctly:
f'(x) = 3(x^2 - 2x) + 9
Next, we need to take half of the coefficient of x, which is -2 in this case, and square it. So, (-2/2)^2 = 1.
f'(x) = 3(x^2 - 2x + 1) + 9 - 3
Simplifying further:
f'(x) = 3(x - 1)^2 + 6
Now, you can observe that the expression (x - 1)^2 is always equal to or greater than 0 for all real values of x. And since 3 multiplied by a non-negative value is always greater than or equal to 0, you can conclude that f'(x) = 3(x - 1)^2 is greater than or equal to 0 for all real values of x. Thus, f(x) is increasing for all real values of x.
2. Show that f(x) = -2x^2 + 3x^2 - 2x + 4 decreases for all real values of x.
For this question, let's find the derivative, f'(x), of the function f(x), which you calculated as:
f'(x) = -6(x - 1/2)^2 - 1/2
This expression is almost in the correct form, but you made a mistake when completing the square. Here's the correct step:
f'(x) = -6(x^2 - x/2) - 1/2
Taking half of the coefficient of x, which is -1/2, and squaring it, we get:
(-1/2)^2 = 1/4
Now, let's adjust the expression:
f'(x) = -6(x^2 - x/2 + 1/4) - 1/2 - 6(1/4)
Simplifying further:
f'(x) = -6(x - 1/2)^2 - 1/2 + 3/2
f'(x) = -6(x - 1/2)^2 + 1
Now, you can observe that the expression (x - 1/2)^2 is always equal to or greater than 0 for all real values of x. And since -6 multiplied by a non-negative value is always less than or equal to 0, you can conclude that f'(x) = -6(x - 1/2)^2 + 1 is less than 0 for all real values of x. Thus, f(x) is decreasing for all real values of x.
Remember, completing the square involves taking half of the coefficient of x, squaring it, and adjusting the expression accordingly. Double-checking your calculations and simplifications will help you avoid errors in future problem-solving.