What are the local extrema, if any, of #f(x)= (4x-3)^2-(x-4)/x #?
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.
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,
Set this equal to zero:
Set this equal to zero: there are no solutions in
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
(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:
(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
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
(Socratic doesn't seem to offer a minus-plus opposite of plus-minus, so we have to write it this way)
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
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
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.