# Video: Finding the Local Maximum and Minimum Values of a Function Involving a Logarithmic Function If Any

Find, if any, the local maxima and minima for 𝑓(𝑥) = 3𝑥² − 2𝑥 − 4 ln 𝑥.

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### Video Transcript

Find, if any, the local maxima and minima for the function 𝑓 of 𝑥 equals three 𝑥 squared minus two 𝑥 minus four times the natural logarithm of 𝑥.

Local maxima and minima are examples of critical points. And we know that for a function 𝑓 of 𝑥, its critical points occur when its first derivative, 𝑓 prime of 𝑥, is equal to zero or is undefined. We therefore need to find an expression for the first derivative of this function. To differentiate each of the first two terms, we can recall the power rule of differentiation. The derivative of three 𝑥 squared is equal to three multiplied by two 𝑥. And the derivative of negative two 𝑥 is equal to negative two. We then need to recall how to differentiate a natural logarithm. Well, we should remember that the derivative with respect to 𝑥 of the natural logarithm of 𝑥 is equal to one over 𝑥.

So, the derivative of negative four times the natural logarithm of 𝑥 is negative four multiplied by one over 𝑥 or negative four over 𝑥. We have our expression for the first derivative then; 𝑓 prime of 𝑥 is equal to six 𝑥 minus two minus four over 𝑥. We then set this expression equal to zero, giving six 𝑥 minus two minus four over 𝑥 equals zero. And we’ll solve the resulting equation. To eliminate the fraction, we can multiply both sides of the equation by 𝑥, giving six 𝑥 squared minus two 𝑥 minus four equals zero, a quadratic equation in 𝑥. We can simplify by dividing by two to give three 𝑥 squared minus 𝑥 minus two equals zero. And then we’ll see if this quadratic can be factored. In fact, it can. And as the coefficient of 𝑥 squared and the constant term are both prime numbers, we can do this using trial and error.

We find that our quadratic factors are three 𝑥 plus two multiplied by 𝑥 minus one. We then take each factor in turn and set it equal to zero and then solve the resulting linear equation. Our first equation, three 𝑥 plus two equals zero, leads to 𝑥 equals negative two-thirds. And our second equation, 𝑥 minus one equals zero, leads to 𝑥 equals one. So, we find two 𝑥-values at which this function 𝑓 of 𝑥 has critical points. Remember, though, that we said that critical points occur when the first derivative is equal to zero or does not exist. So, we also need to check if there are any values of 𝑥 for which our first derivative will be undefined. Well, as we have this factor of negative four over 𝑥 in our definition of 𝑓 prime of 𝑥, 𝑓 prime of 𝑥 does not exist when 𝑥 equals zero because division by zero is undefined.

However, if we look back at our original function 𝑓 of 𝑥, we see that it includes the term negative four times the natural logarithm of 𝑥. And logarithms are only defined for positive values of 𝑥. So, the domain of our original function 𝑓 of 𝑥 is 𝑥 greater than zero. This value, 𝑥 equals zero, is not included in the domain of our original function. And therefore, we don’t need to be concerned that 𝑓 prime of 𝑥 is undefined for this value of 𝑥. For this same reason, we see that whilst 𝑥 equals negative two-thirds is a perfectly valid solution to the quadratic equation, it isn’t a valid solution when we’re talking about critical points because this 𝑥-value isn’t in the domain of our original function. So, in fact, we find that there is only one 𝑥-value at which our function 𝑓 of 𝑥 has a critical point. It’s when 𝑥 is equal to one.

Next, we need to evaluate the function itself at this critical point. So, we substitute 𝑥 equals one into our definition of 𝑓 of 𝑥, giving three multiplied by one squared minus two multiplied by one minus four multiplied by the natural logarithm of one. Now, we should recall that the natural logarithm of one is just zero. So, this is equal to three minus two, which is equal to one. We know then that our function 𝑓 of 𝑥 has one critical point, at the point with coordinates one, one. But we don’t yet know whether it is a local minimum or local maximum. In order to determine this, we need to consider the shape of the function around this point, which we can do by performing the first derivative test.

We know that the first derivative of our function is equal to zero at the critical point. That’s when 𝑥 is equal to one. We then choose values of 𝑥 either side of our critical point and consider the sign of the first derivative for each of these. Now usually, we would try to choose the integer values immediately either side of our critical points, so that would be zero and two here. But as our function is only defined for 𝑥 strictly greater than zero, I’ve chosen 𝑥 equals a half to be the 𝑥-value to the left of our critical point. When 𝑥 equals one-half, the first derivative 𝑓 prime of 𝑥 is equal to six multiplied by a half minus two minus four over a half. That’s three minus two minus four times two which is eight, which is equal to negative seven. And so, we find that the first derivative will be negative when 𝑥 is equal to one-half. That’s to the left of our critical point.

When 𝑥 equals two, 𝑓 prime of 𝑥 would be equal to six multiplied by two minus two minus four over two. That’s 12 minus two minus two, which is equal to eight. And so, we find that the first derivative of our function is positive to the right of our critical point. We then consider the shape of the curve. Travelling from left to right, the first derivative changes from negative to zero to positive. And so, we see that this critical point is a local minimum. We can conclude then that our function 𝑓 of 𝑥 has a local minimum which is equal to one, and this occurs when 𝑥 is also equal to one.