Video: Finding the Local Maximum and Minimum Values of a Polynomial Function

Find the local maxima/minima of the function 𝑓(π‘₯) = 3π‘₯⁴ βˆ’ 2π‘₯Β³.

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Video Transcript

Find the local maxima or minima of the function 𝑓 of π‘₯ equals three π‘₯ to the fourth power minus two π‘₯ cubed.

Local maxima and minima are examples of critical points for function. And we know that the critical points for function occur when its first derivative, 𝑓 prime of π‘₯, is equal to zero or is undefined. Looking at our function 𝑓 of π‘₯, we see that it is a polynomial. And we know that the derivative of a polynomial is defined for all values in its domain. So we don’t need to be concerned about 𝑓 prime of π‘₯ being undefined. We only need to look for 𝑓 prime of π‘₯ equal to zero to determine critical points.

Let’s first find an expression for 𝑓 prime of π‘₯ then. And we can do this using the power rule of differentiation. 𝑓 prime of π‘₯ is equal to three multiplied by four π‘₯ cubed. Remember, we bring the exponent down and then reduce the exponent by one minus two multiplied by three π‘₯ squared, which simplifies to 12π‘₯ cubed minus six π‘₯ squared. To find the π‘₯-coordinates of critical points, we then set our first derivative equal to zero and solve the resulting equation for π‘₯.

Now, we can take a common factor of six π‘₯ squared from each of these terms, giving six π‘₯ squared multiplied by two π‘₯ minus one is equal to zero. Six is not equal to zero. So setting the remaining two factors in turn equal to zero, we have π‘₯ squared equals zero, leading to π‘₯ equals zero and two π‘₯ minus one equals zero, leading to π‘₯ equals one-half. We’ve therefore found that this function 𝑓 of π‘₯ has critical points at π‘₯ equals zero and π‘₯ equals one-half.

We then need to evaluate the function itself at each of the critical points. Firstly, substituting π‘₯ equals zero gives 𝑓 of zero equals three multiplied by zero to the fourth power minus two multiplied by zero cubed, which is equal to zero. Substituting π‘₯ equals one-half gives three multiplied by one-half to the fourth power minus two multiplied by one-half cubed. That’s three over 16 minus one over four or one-quarter, which we can think of as three over 16 minus four over 16, which is negative one over 16. The critical points of this function then occur at zero, zero and one-half, negative one over 16. But we don’t yet know what type of critical points there are. There’re three possibilities. They could be local maxima, local minima, or points of inflection.

In order to classify each of our critical points, we need to use the second derivative test. The sign of the second derivative of our function tells us which type of critical point we have. If the second derivative of our function is negative, then this means that the first derivative or slope of the function is decreasing. And so we have a local maxima. If the second derivative is positive, then the slope or first derivative of the function 𝑓 prime of π‘₯ is increasing. And so we have a local minimum. If the second derivative is equal to zero, then the critical point could be a point of inflection. But this isn’t sufficient. It is possible to have a local minimum or a local maximum when the second derivative is equal to zero. So if we find the second derivative is equal to zero, we need to perform other checks in order to classify the critical point.

Let’s begin then by finding an expression for the second derivative of our function. And to do this, we need to differentiate the expression we found for the first derivative, which remember was 12π‘₯ cubed minus six π‘₯ squared. Differentiating gives 12 multiplied by three π‘₯ squared minus six multiplied by two π‘₯. That simplifies to 36π‘₯ squared minus 12π‘₯. Let’s check the critical point when π‘₯ is a half first of all. Substituting π‘₯ equals one-half into our second derivative gives 36 multiplied by one-half squared minus 12 multiplied by one-half. That’s 36 over four minus 12 over two or nine minus six, which is equal to three. We’re not particularly interested in the value of the second derivative other than to notice that it is greater than zero. And so, by the second derivative test, we find that this critical point is a local minimum.

When we substitute π‘₯ equals zero for our other critical point, we find that the second derivative is also equal to zero. And so here, the second derivative test hasn’t helped us with classifying this critical point. We need to perform a different check. And we’ll use the first derivative test. What we’ll do is check the sign of the first derivative either side of our critical point. So we can consider the shape of the curve around the critical point.

Now, as our next critical point occurs when π‘₯ equals one-half, we need to choose π‘₯-values quite close to zero. So we’ll choose π‘₯-values of negative one-quarter and one-quarter. We already know that 𝑓 prime of π‘₯, the first derivative, is equal to zero when π‘₯ is equal to zero, at the critical point itself. Substituting π‘₯ equals negative one-quarter into our first derivative, which remember was 12π‘₯ cubed minus six π‘₯ squared, we find that the first derivative is equal to negative nine over 16 at this point. We’re not particularly interested in the actual value. But we are interested in the fact that it is a negative value.

Substituting π‘₯ equals positive a quarter into our first derivative gives 12 multiplied by one-quarter cubed minus six multiplied by one-quarter squared, which is negative three over 16. And again, what we’re interested in is the fact that the first derivative is negative here. This helps us to picture the shape of the curve at this critical point. If the slope is negative, then zero, then negative again, we can sketch the shape. And we see that this critical point is in fact a point of inflection. We only need to perform this first derivative test when the second derivative test fails to distinguish between a local minimum, a local maximum, or a point of inflection. But actually, the first derivative test is an acceptable method in its own right.

The question only asked us about local maxima or minima. So we don’t need to mention the point of inflection in our final answer although we did need to check it in our working. We can conclude then that the function 𝑓 of π‘₯ has a local minimum point at one-half, negative one sixteenth.

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