# 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.

- Standard approach points us in a laborious direction

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://tutor.hix.ai - 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.

- Solve a simplified problem

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

- 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://tutor.hix.ai).

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...

- 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):

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

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.

By signing up, you agree to our Terms of Service and Privacy Policy

To find the local extrema of ( f(x) = (4x - 3)^2 - \frac{x - 4}{x} ), we first need to find the critical points by setting the derivative equal to zero and solving for ( x ).

First, find the derivative of ( f(x) ):

[ f'(x) = 2(4x - 3) \cdot 4 - \left(1 - \frac{4}{x^2}\right) ]

Now, set ( f'(x) = 0 ) and solve for ( x ):

[ 8(4x - 3) - 1 + \frac{4}{x^2} = 0 ]

[ 32x - 24 - 1 + \frac{4}{x^2} = 0 ]

[ 32x - 25 + \frac{4}{x^2} = 0 ]

[ 32x^3 - 25x^2 + 4 = 0 ]

[ (4x - 1)(8x^2 - 17x - 4) = 0 ]

This gives us three potential critical points: ( x = \frac{1}{4} ) and the solutions to ( 8x^2 - 17x - 4 = 0 ).

Solve ( 8x^2 - 17x - 4 = 0 ) using the quadratic formula:

[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

[ x = \frac{17 \pm \sqrt{289 + 128}}{16} ]

[ x = \frac{17 \pm \sqrt{417}}{16} ]

[ x = \frac{17 \pm \sqrt{3 \cdot 139}}{16} ]

[ x = \frac{17 \pm \sqrt{3} \sqrt{139}}{16} ]

Therefore, the critical points are:

[ x = \frac{1}{4}, \quad x = \frac{17 + \sqrt{3} \sqrt{139}}{16}, \quad x = \frac{17 - \sqrt{3} \sqrt{139}}{16} ]

Next, evaluate ( f(x) ) at these critical points and at endpoints if the interval is bounded to determine the local extrema. However, since there's no specified interval, we'll only analyze the critical points:

[ f\left(\frac{1}{4}\right) = \left(4\left(\frac{1}{4}\right) - 3\right)^2 - \frac{\frac{1}{4} - 4}{\frac{1}{4}} = 0 ]

[ f\left(\frac{17 + \sqrt{3} \sqrt{139}}{16}\right) = \left(4\left(\frac{17 + \sqrt{3} \sqrt{139}}{16}\right) - 3\right)^2 - \frac{\frac{17 + \sqrt{3} \sqrt{139}}{16} - 4}{\frac{17 + \sqrt{3} \sqrt{139}}{16}} \approx -24.537 ]

[ f\left(\frac{17 - \sqrt{3} \sqrt{139}}{16}\right) = \left(4\left(\frac{17 - \sqrt{3} \sqrt{139}}{16}\right) - 3\right)^2 - \frac{\frac{17 - \sqrt{3} \sqrt{139}}{16} - 4}{\frac{17 - \sqrt{3} \sqrt{139}}{16}} \approx 2.154 ]

Thus, the local minimum is approximately ( 2.154 ) and the local maximum is approximately ( -24.537 ).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How to sketch graph of f(x)= 1/(x²-4x)?
- #f(x)=-1/2x^2+6x-13#. The line #y=1/2x+c# is orthogonal to the graph of #f#. Then what is #c#?
- What are the extrema and saddle points of #f(x, y) = 6 sin x sin y# on the interval #x,y in[-pi,pi]# ?
- Given the function #f(x)=(x-3)^(2/3)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,4] and find the c?
- How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f(x)=x^3+5x^2#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7