# What are the local extrema, if any, of #f(x)= (4x-3)^2-(x-4)/x #?

##### 1 Answer

#### Answer:

The only extremum is

But you have to solve a cubic equation to get there and the answer is not at all 'nice' - are you sure the question is correctly typed in? I have also included suggestions for how to approach the answer without going into the amount of analysis shown fully below.

#### Explanation:

**1. Standard approach points us in a laborious direction**

First calculate the derivative:

so (by chain and quotient rules)

Then set this equal to 0 and solve for

We have a cubic equation, which is solvable by radicals, but this is far from an easy process. We know that this equation will in general have three roots, but not that they will all be real, although at least one of them will be - that at least one will be we know from the Intermediate Value Theorem - https://en.wikipedia.org/wiki/Intermediate_value_theorem - which tells us that because the function goes to infinity at one end and minus infinity at the other, then it must take all values in between at one point or another.

Trialling a few simple values (1 is often an informative and quick value to try), we see that there is a root somewhere between 1/2 and 1, but we don't find any obvious solutions to simplify the equation with. Solving a cubic equation is a long and tedious process (which we'll do below), so it is worth trying to inform one's intuition before doing so. Trialling solutions further, we find that it is between 0.9 and 0.91.

**2. Solve a simplified problem**

The function consists of the difference of two terms,

**First term, #f_1#**

Set this equal to zero:

**Second term, #f_2#**

Set this equal to zero: there are no solutions in

Note that

So now we are certain that we are only looking for one solution but we do not have a good answer for it.

**3. Numerically approximate the answer**

In professional situations requiring the solution of these kinds of problems, often the quickest way to get to where you need to get is to perform a numerical approximation. A pretty good one for finding roots of a function is the Newton-Raphson method (https://en.wikipedia.org/wiki/Newton%27s_method).

Which is: to find a root of a function

Recall our function and its derivative:

So we might guess 0.5 as our root, making

So we can find the answer with arbitrary precision, but the full answer requires an analytic solution, something that we noted above would be hard. So here we go...

**4. Solve the full problem, slowly and painfully**

Now let's do the full cubic solution (you're going to have to love algebra to solve this one properly):

**First, divide through to make the leading term have coefficient 1** :

**Second, make the following substitution to the variable #y# to remove the #x^2# term:**

Substitute

(Expand brackets, remembering the Binomial theorem: https://en.wikipedia.org/wiki/Binomial_theorem)

(Notice that the two

We now have the same number of terms as we did before, because we previously had no

**Third, make another substitution (Vieta's substitution: http://mathworld.wolfram.com/VietasSubstitution.html) to turn this into a quadratic:**

Substitute

(Notice that both

(Now, you may well ask what on earth the benefit of this is - we've fiddled with our degree 3 equation until we have a degree 6 equation, surely a loss... But we can now think of it as a quadratic equation in

**Fourth, solve the quadratic equation for #w^3#**

Using the quadratic equation:

We have an answer! Now we just have to relate it back to our original variable

**Fifth, convert back to our original terms**

Take the cube root:

Recall how we related

Now

(Socratic doesn't seem to offer a minus-plus opposite of plus-minus, so we have to write it this way)

Thus

If we multiply out the minus signs in the second large term we can see that we obtain two identical expressions, so we can drop the quadratic plus/minus signs, and simplify to

Finally (!) recall that we set

Thus

**Sixth, deduce how many of these roots are real**

The two expressions in the cube roots each have one real root and two conjugate imaginary roots. A real number

**Conclusion**

Therefore there is only one real root for

or, in decimal

We can deduce that this is a minimum of the function by the fact that there is only one extremum and the function tends to positive infinity at both ends.