The smallest slope of f(x)=6x^2-x^3 for 0 is less than or equal to x is less than or equal to 6 occurs at x=
A. 0
B. 2
C. 3
D. 4
E. 6
I got E
You're asking for
minimum value of f'(x) for 0≤x≤6 where f(x)=6x²-x³.
First we need to know if f'(x) is increasing, decreasing, or there's a minimum within [0,6].
We can find out by first finding
f'(x)=12x-3x²
Find critical points for f'(x):
f"(x)=12-6x
so f"(x)=0 at x=2.
So check f'(x) for x=0,2,6 and locate the minimum of f'(x):
f'(0)=0
f'(2)=12
f'(6)=-36
so f'(6) is least, so choice E is correct.
Hmm, let me calculate it for you in a "clowny" way!
Well, the function f(x) = 6x^2 - x^3 is kind of like a roller coaster ride. And we all know that roller coasters have their ups and downs, right?
So, let's imagine you're riding this math coaster from x = 0 to x = 6. As you start at x = 0, you're at the very bottom of the hill, so the slope is practically zero.
But then, as you start climbing up, the slope increases and reaches its peak somewhere along the ride. And according to my calculations, that peak occurs when x is E. Yes, at x = 6!
So, the answer is E, my dear friend. You'll experience the smallest slope at the very end of the roller coaster ride. Enjoy the mathematical thrill!
To find the smallest slope of the function f(x) = 6x^2 - x^3 in the given interval 0 ≤ x ≤ 6, we need to determine the critical points of the function.
Step 1: Find the derivative of f(x) with respect to x.
f'(x) = 12x - 3x^2
Step 2: Set f'(x) equal to zero and solve for x to find the critical points.
12x - 3x^2 = 0
3x(4 - x) = 0
Setting each factor equal to zero, we have:
3x = 0 or 4 - x = 0
Solving the equations, we find:
x = 0 or x = 4
Step 3: Check the second derivative to determine if the critical points are at local maximum or minimum.
Taking the second derivative:
f''(x) = 12 - 6x
Since f''(x) = 0 does not have any solutions in the given interval, we can proceed to step 4.
Step 4: Evaluate the function at the critical points and the endpoints of the interval.
f(0) = 6(0)^2 - (0)^3 = 0
f(4) = 6(4)^2 - (4)^3 = 96 - 64 = 32
Step 5: Compare the values to find the smallest slope.
Since f(0) = 0 and f(4) = 32, the smallest slope occurs at x = 0.
Therefore, the answer is A. 0.
To find the smallest slope of the function f(x) = 6x^2 - x^3 for 0 ≤ x ≤ 6, we need to calculate the derivative of the function and then find the critical points.
First, let's find the derivative of f(x) with respect to x:
f'(x) = d/dx(6x^2 - x^3)
f'(x) = 12x - 3x^2
Now, we need to find the critical points by setting the derivative equal to zero and solving for x:
12x - 3x^2 = 0
Factoring out an x, we get:
x(12 - 3x) = 0
Setting each factor equal to zero:
x = 0
12 - 3x = 0
For the first equation, x = 0 is already within the given interval, so it is a potential critical point.
For the second equation, solving for x, we get:
12 - 3x = 0
3x = 12
x = 4
Now we have two potential critical points: x = 0 and x = 4.
To find the smallest slope, we need to evaluate the slopes on the interval [0, 6] and compare the values.
Slope at x = 0: f'(0) = 12(0) - 3(0)^2 = 0
Slope at x = 4: f'(4) = 12(4) - 3(4)^2 = 48 - 48 = 0
Based on the calculations, both potential critical points have a slope of 0. However, we need to consider the fact that x = 0 is already within the given interval, so the smallest slope occurs at x = 0.
Therefore, the answer is A. x = 0.