Is it possible to solve and explain this question in a different way? I am working on an online mathematics Helper web site and I need solutions in different ways. This is not a homework or any kind of exam as I already have a solution of these questions. I just need different ways of solutions so that I can write different algorithms on those solutions and then call them in software coding randomly.
Were you needing alternatives for the other solutions? The intercepts should be easy.
For the min/max you can do the second derivative of the original function at the vertex. If it is negative then
A very different approach (that may be much easier to program is:
Take the first derivative of the function
Solve it for 0 (i.e. what x gives you a first derivative =0?)
Plug that x in the original function to get the value of y
Were you needing alternatives for the other solutions? The intercepts should be easy.
For the min/max you can do the second derivative of the original function at the vertex. If it is negative then the vertex is a maximum, a positive indicates a minimum, and 0 indicates (I forget the proper name) that it is a "flat spot with one side increasing and the other decreasing".
0
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Can you please show me a little bit work as I am not in touch with calculus since a long time.
I think first derivative of that equation would be:
2x-8+0
2x=8
x=4
The original function is:
x^2-8x+7=0
substituting x
8-32+7=0
y=31
I am not sure if this is correct. or I may mixing the integration with derivation.
The second derivative of x^2-8x+7 is the first derivative of the first derivative.
So... the first derivative is 2x-8
The second derivative is 2.
Since the second derivative is positive, then the point is a minimum. That is, the function slopes upward on either side of that point.
Note that the second derivative doesn't depend on x. That means that as you move from left to right on the graph, its slope increases. When the slope is negative (as in to the left of the vertex), "slope increases" means it becomes less negative.
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