WizardLMTeam/WizardMath-7B-V1.1 · results are quite good

I have run some tests using ChatLLM.cpp. The results are quite good.

Some examples:

You > find the peak of f(x) = 3 (x - 1)^2 - 5
A.I. >
[Rewritten Answer]: To find the peak of the function f(x) = 3(x - 1)^2 - 5, we need to follow these steps:

Step 1: Identify the quadratic function
The given function is a quadratic function in the form of f(x) = ax^2 + bx + c, where a = 3, b = -6, and c = -5.

Step 2: Find the vertex of the parabola
The vertex of a parabola in the form of f(x) = ax^2 + bx + c is given by the formula:

x_vertex = -b / (2a)

Step 3: Plug in the values of a and b
Now, we will plug in the values of a and b that we found earlier:

x_vertex = -(-6) / (2 * 3)

Step 4: Simplify the expression
Simplify the expression to find the x-coordinate of the vertex:

x_vertex = 6 / 6
x_vertex = 1

Step 5: Find the y-coordinate of the vertex
Now that we have the x-coordinate of the vertex, we can find the y-coordinate by plugging it back into the function:

f(1) = 3(1 - 1)^2 - 5

Step 6: Simplify the expression
Simplify the expression to find the y-coordinate of the vertex:

f(1) = 3(0)^2 - 5
f(1) = -5

Step 7: State the final answer
The peak of the function f(x) = 3(x - 1)^2 - 5 is at the point (1, -5).

The answer is: the peak of the function f(x) = 3(x - 1)^2 - 5 is at the point (1, -5).
You > is it the maximum or minimum of the function
A.I. > The peak of a function is the highest or lowest point of the function. In this case, since the coefficient of the squared term (x - 1)^2 is positive (3), the parabola opens upwards, which means it has a minimum point. Therefore, the peak of the function f(x) = 3(x - 1)^2 - 5 is the minimum of the function.