Derivatives of linear inequalities; for example, x^2 - 4x + 3 < 0

[Thermo]

Hi all,

I wondered if there is a method for finding the derivatives of linear inequalities; e.g.
x^2 - 4x + 3 < 0

I don't think the answer is simply: 2x - 4 < 0

Because if a function is increasing, you cannot expect that its derivative will increase too.

I'm a bit stuck.

Many thanks
Nick

 

[Otis]

method for finding the derivatives of linear inequalities

eg: x^2 - 4x + 3 < 0

Hi Nick. That's not a linear polynomial (it's quadratic), and I'm not sure what you mean by talking about derivatives of inequalities. We take derivatives of functions, not of equations or inequalities.

I'll post some basic statements below, yet I suspect that either you or I have misinterpreted something. Please reply by clarifying what you're trying to do or understand.

I don't think the answer is simply: 2x - 4 < 0

Okay, but what is the question?

If the exercise asks you to solve the quadratic inequality, then the solution is a set of x-values. Specifically, it's the set of all inputs that cause the quadratic function to output a negative value.

The leading coefficient of the function x^2 – 4x + 3 is positive, so we know the parabolic graph opens upward.

The discriminant (b^2–4ac) of the function is 4, so we know the graph has two x-intercepts.

Therefore, we know the graph lies below y=0 between the x-intercepts, so the solution to the quadratic inequality is the set of all x-values strictly inbetween those two roots.

---

The quadratic polynomial's first derivative is the linear function 2x–4.

2x – 4 < 0

The solution to the linear inequality is also a set of x-values. That set contains all values of x that cause the function 2x–4 to be negative.

---

… if a function is increasing, you cannot expect that its derivative will increase too.

That is correct, but we do expect that the derivative will be positive anywhere the function is increasing.

---

In summary, the first derivative of a function is new function that outputs the slope (i.e., the rate of change) of the original function. Let's name the quadratic function f. By convention, the name of the first derivative function is f' (f-prime).

f(x) = x^2 – 4x + 3

f'(x) = 2x – 4

What is the slope of the parabolic graph when x is -1? That slope is f'(-1). In other words, the slope is -6.

What is function f's rate of change when x=2? That number is f'(2)=0.

What's the slope at x=7/2? It's f'(7/2) which is 3.

Here is a graph of f(x) and f'(x) for x-values from -1 to 5. Again, function f' gives the slope of function f, at each x in the domain.

 

[Thermo]

Hi Otis,

I appreciate your extensive reply.
Nothing has been misinterpreted, although I used the word linear, (oversight), for the inequality.
It was a mock test question from an exam paper, see below:



I think that adding a constant k to the inequality, equating to zero and then solving as usual would be sufficient?

Many thanks again

 

[Otis]

It was a mock test question

Okay, Nick. I'm going to mock the question a bit, then.

First of all, it's simply not correct to call (d) an "equation". It is an inequality.

Yet, I understand what they're asking for overall. They want the derivative of the function that appears in each part of the exercise.

And, when they say, "Find the equations of the gradient, [imath]\frac{dy}{dx}[/imath]…", they mean your answer must take the form of an equation, like this:

[imath]\frac{dy}{dx}[/imath] = < write the function's derivative here >

In other words, you're not finding derivatives "of the" equations. You're just reporting each answer in the form of an equation.

Therefore, you find the derivative in (d) as usual, and write the answer as [imath]\frac{dy}{dx}=2x-4[/imath].

If they're trying to make some point by presenting (d) as an inequality instead of just writing y=x^2–4x+3, then that point is lost on me.

 

[Otis]

I think that adding a constant k to the inequality, equating to zero and then solving as usual would be sufficient?

I'm not sure what you're thinking, by adding k. On which side of the given inequality would you add k?

When you say, "solving as usual", are you talking about a method to find the derivative of x^2–4x+3?

If so, then what is that method? Or, instead of describing the steps you have in mind, you could post a picture of your attempt on part (d).

 

[Thermo]

Thanks Otis, I agree it is an inequality, not and equation and, I also can't see why they used that. Anyway, I feel better that it's probably a poorly thought-out/worded question.
Cheers