Please, refer to this link.
Can anyone explain how was that decision taken that the minimum is at about $0.8685$?
Understanding the 1st derivative test in optimization.
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derivatives
optimization
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2When the first derivative is zero, we have a saddle point. If the first derivative goes from negative to positive, we know the function went from decreasing to increasing, so the saddle point is a minimum. – 2017-01-07
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0@Kaynex You are confusing stationary (or critical) points with saddle points. – 2017-01-07
1 Answers
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First, let's plot the function in question so we can get an idea of the shape of the curve over the range of interest for this problem.
Next, let's find the critical points by finding the first derivative, equating it to zero and finding the roots.
$$f(x) = (x-2)^2 + x^3 \implies f'(x) = 3 x^2+2 x-4=0$$
The roots are $$\implies x = \dfrac {1} {3}\left (-\sqrt{13} - 1 \right) \approx - 1.5352, x = \frac {1} {3}\left (\sqrt{13} - 1 \right) \approx 0.86852$$
Next, we will make use of the First Derivative test, by summarizing all of this information in two pictures, the left critical point and the right critial point.
If we look at the sign of the slope a bit to the left ($x = -1.5352 - \epsilon$), we have a positive slope, which tells us the graph is increasing. The sign of the slope is zero at the critical point. If we look at the sign of the slope a bit to the right ($x = -1.5352 + \epsilon$), we have a negative slope, which tells us the graph is decreasing. Taking all of this data together and summarizing it in the first figure below, we can tell that we have a local maximum by looking at the shape of the graph information.
We can then repeat this process for the other critical point, $x = 0.8685$ (see second figure below) and conclude that that is the local minimum.
Compare the figure above with these two figures and make sure it all makes sense. Next, we can make use of the Second Derivative test, but that was not used in this problem.
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