# Homework Help

[stargazer]

I'm struggling with a problem set for my Calculus class. We're covering optimization techniques, and I need help on question 5 from the assignment. It asks to find the minimum of the function f(x) = 3x^2 - 6x + 9 over the interval [0, 4]. I thought I could just plug in the endpoints and find any critical points within the interval to use as a candidate solution, but my calculations aren't matching the given answer. Can someone else help me understand where this went wrong?

 

[wisdom]

To find the minimum of f = 3x^2 - 6x + 9 over [0, 4], you're correct about considering endpoints and critical points.

First, let's evaluate f and f:

f = 3*0^2 - 6*0 + 9 = 9

f = 3*4^2 - 6*4 + 9 = 3*16 - 24 + 9 = 48 - 24 + 9 = 33

Next, we'll find the critical points. The derivative of f is f' = 6x - 6.

Set the derivative equal to zero:

6x - 6 = 0
=> 6x = 6
=> x = 1

Now that you have the endpoints and the critical point , check which one gives the minimum value. Since f is a quadratic function, it has a parabolic shape that opens upwards due to its positive leading coefficient . Therefore, the minimum occurs at a vertex or critical point when considering any open interval.

To verify which one yields the smallest value, compare f, f, and f:

f = 3*1^2 - 6*1 + 9 = 3 - 6 + 9 = 6
Comparing these results, you'll see that f = 6 is the smallest value among those three.

Therefore, your approach was sound and just required some careful attention to details in verifying critical points.

 

[musical]

# Homework Help

You are correct about considering endpoints and critical points for finding the minimum of f = 3x^2 - 6x + 9 over [0, 4].

First, let's evaluate f and f:

f = 3*0^2 - 6*0 + 9 = 9

f = 3*4^2 - 6*4 + 9 = 3*16 - 24 + 9 = 48 - 24 + 9 = 33

Next, we'll find the critical points. The derivative of f is f' = 6x - 6.

Set the derivative equal to zero:

6x - 6 = 0
=> 6x = 6
=> x = 1

Now that you have the endpoints and the critical point , check which one gives the minimum value. Since f is a quadratic function, it has a parabolic shape that opens upwards due to its positive leading coefficient. Therefore, the minimum occurs at a vertex or critical point when considering any open interval.

To verify which one yields the smallest value, compare f, f, and f:

f = 9
f = 3*1^2 - 6*1 + 9 = 3 - 6 + 9 = 6
f = 33

Comparing these results, you'll see that f=6 is the smallest value among those three.

Therefore, your approach was sound and just required some careful attention to details in verifying critical points..