Inverse functions

What Nour says

1. 1. Continuous and strictly monotonic functions on an interval

   Today, the whole chapter: "Inverse functions", Year 4, Mathematics section.

   We will build this step by step, from monotonic functions to drawing the inverse curve.

   Let's start with continuous and strictly monotonic functions on an interval.

   A function is continuous on an interval when its curve has no break, no jump, no hole.

   You already saw continuity in an earlier chapter, so this is just a quick reminder.

   A function is strictly increasing on an interval when it always goes up as x goes up.

   In other words, for any a and b in the interval, if a is less than b then f(a) is less than f(b).

   A function is strictly decreasing when it always goes down as x goes up.

   Here if a is less than b then f(a) is greater than f(b).

   A function that is always increasing or always decreasing on an interval is called strictly monotonic there.

   This phrase, a function continuous and strictly monotonic on an interval, is a key condition in this chapter.

   It means the curve is unbroken and it never turns back, it keeps one single direction.

   Here is why this condition matters so much for what comes next.

   When a curve always goes the same direction, each height on the curve comes from exactly one x.

   No horizontal line can cross it twice, because the function never comes back down, or never comes back up.

   This single-crossing property is exactly what lets us build an inverse function later in this chapter.

   Now let's look at a worked example, as you would see in a bac exam.

   Let f be the function defined on the interval from 0 to plus infinity by f of x equals x squared plus 1.

   We must show that f is continuous and strictly increasing on this interval.

   First, f is a polynomial function, so it is continuous everywhere, in particular on this interval.

   Second, we check f is strictly increasing by comparing two values a and b with 0 less than a less than b.

   Since a and b are positive and a is less than b, squaring keeps the order, so a squared is less than b squared.

   Adding 1 to both sides keeps the order, so f of a is less than f of b.

   This proves f is strictly increasing on the interval from 0 to plus infinity.

   So f is continuous and strictly increasing on this interval, our key condition is satisfied.

   Let's check with a point: at x equals 2, f of 2 equals 5.

   At x equals 3, f of 3 equals 10, which is bigger, as expected since f increases.

   Notice the curve only rises, it never comes back down, matching our proof.

   So the method is always the same: check continuity first, then compare f of a and f of b.

   This condition, continuous and strictly monotonic on an interval, is the starting point of the whole chapter.

   Keep this picture in mind: an unbroken curve going only one way, never turning back.

   Next we will use this exact condition to build something new, called an inverse function.

   What does it mean for a function to be strictly monotonic on an interval?

2. 2. The bijection theorem

   Now let's move to the bijection theorem.

   This theorem says when a function has an inverse function.

   If f is continuous and strictly monotonic on an interval I, then f gives a bijection from I onto f of I.

   That is, every value in f of I comes from exactly one x in I, no more, no less.

   This also means f has an inverse function, written f to the power minus 1.

   The inverse function sends each y back to the unique x that gave it.

   So if f of a equals b, then f inverse of b equals a.

   Here is why we need both conditions, continuity and strict monotonicity, together.

   Without continuity, the curve could jump and skip values, leaving gaps in f of I.

   Without strict monotonicity, the curve could go up then down, hitting the same value twice.

   Both conditions together guarantee one output for one input, and one input for one output.

   This is exactly the single-crossing idea we saw with the horizontal line in the last section.

   Now let's apply this with a worked example from a bac-style exercise.

   Let f be defined on the interval from 0 to plus infinity by f of x equals square root of x.

   We want to show f realizes a bijection from this interval onto an interval we must find.

   First, f is continuous on this interval, since the square root function is continuous there.

   Second, f is strictly increasing, because larger x gives a larger square root.

   So by the bijection theorem, f is a bijection from the interval from 0 to plus infinity, onto f of that interval.

   To find f of the interval, we look at the limits of f at the two endpoints, 0 and plus infinity.

   Since f is continuous and strictly increasing, f of the interval is the interval from f of 0 to the limit.

   So f is a bijection from the interval from 0 to plus infinity, onto itself.

   This means f has an inverse function, also defined on the interval from 0 to plus infinity.

   Let's check with a point: f of 4 equals 2, since the square root of 4 is 2.

   So f inverse of 2 must equal 4, going back the other way.

   This matches the idea: f and f inverse undo each other exactly.

   Notice the method: check continuity, check strict monotonicity, then find the image interval using limits.

   This method works for any function you meet in this type of exercise.

   The bijection theorem is the tool that guarantees an inverse function exists.

   Without it, we could never be sure f inverse is well defined.

   Next, we will study the properties this inverse function always has.

   Why does continuity plus strict monotonicity guarantee each value is reached exactly once?

3. 3. Properties of the inverse function

   Let's now look at the properties of the inverse function.

   First, f inverse has the same sense of variation as f.

   If f is strictly increasing, then f inverse is also strictly increasing on f of I.

   If f is strictly decreasing, then f inverse is also strictly decreasing.

   Second, applying f then f inverse, or f inverse then f, brings you back to where you started.

   Third, the domain of f inverse is the range of f, and the range of f inverse is the domain of f.

   Here is why these properties always hold, without exception.

   Since f sends each x in I to one unique y, f inverse just reverses that exact link.

   Reversing a link that only goes up keeps it going up, that is why monotonicity is preserved.

   Now let's work through a bac-style example on these properties.

   Let f be defined on the interval from 1 to plus infinity by f of x equals x squared minus 1.

   We accept that f is continuous and strictly increasing, so f has an inverse function on f of I.

   Since f of 1 equals 0, and f tends to plus infinity, f of I is the interval from 0 to plus infinity.

   So f inverse is defined on the interval from 0 to plus infinity, and takes values in the interval from 1 to plus infinity.

   Because f is strictly increasing, we know f inverse is also strictly increasing.

   Let's check the point f of 3 equals 8, so f inverse of 8 must equal 3.

   We verify f inverse of f of 3 gives back 3, confirming the undo property.

   This matches exactly what we wrote: f inverse of f of x equals x.

   So the properties are not just theory, they work on every concrete example like this one.

   Keep in mind these three properties: same variation, the undo rule, and swapped domain and range.

   They always come from the bijection theorem we studied before.

   Next, we will see how to differentiate an inverse function.

   If f is strictly decreasing, what can you say about f inverse?

4. 4. Derivative of an inverse function

   Now let's study the derivative of an inverse function.

   This is a rule that gives the derivative of f inverse, using the derivative of f.

   In other words, we find the slope of f inverse without recomputing everything from scratch.

   Suppose f is continuous, strictly monotonic on I, and differentiable at a point a in I.

   Let b equal f of a, so a equals f inverse of b.

   If the derivative of f at a is not zero, then f inverse is differentiable at b.

   And the derivative of f inverse at b equals one divided by the derivative of f at a.

   Here is why this formula makes sense, looking at the curves.

   The curve of f inverse is the curve of f reflected across the line y equals x.

   Reflecting a curve across this line swaps the roles of rise and run, so the slope gets inverted.

   That is why the derivative of f inverse is one over the derivative of f, not the derivative of f itself.

   If the derivative of f at a were zero, the reflected curve would have a vertical tangent, so no derivative.

   Now let's apply this formula with a worked bac-style example.

   Let f be defined on the interval from 0 to plus infinity by f of x equals x squared plus 1.

   We accept f is a bijection from this interval onto the interval from 1 to plus infinity.

   We want the derivative of f inverse at the point b equals 5.

   First we find a such that f of a equals 5, so a squared plus 1 equals 5.

   Since a must be in the interval from 0 to plus infinity, a equals 2.

   Next, the derivative of f at x is two x, so the derivative of f at 2 is four.

   By the formula, the derivative of f inverse at 5 equals one over the derivative of f at 2.

   This gives the derivative of f inverse at 5 equals one quarter.

   So the slope of f inverse at the point with x equals 5 is exactly one quarter.

   Notice we never built the formula for f inverse itself, we just used f and its derivative.

   This method always works: find a, compute the derivative of f at a, then take the reciprocal.

   It saves time, since finding f inverse directly can be hard or even impossible to write down.

   Keep this formula in mind: the derivative of f inverse at b is one over the derivative of f at a.

   Remember also the condition: it only works when the derivative of f at a is not zero.

   This rule connects the two curves through their reflection across the line y equals x.

   Next, we will use that same reflection to draw the graph of f inverse.

   In the formula for the derivative of f inverse, what condition must the derivative of f at a satisfy?

5. 5. Graph of the inverse function

   Let's now learn how to draw the graph of the inverse function.

   This is about tracing the curve of f inverse directly from the curve of f.

   That is, we get the second curve without computing a new formula at all.

   The curve of f inverse is the reflection of the curve of f across the line y equals x.

   Here is the curve of f, strictly increasing on its interval.

   Every point a, b on the curve of f gives a point b, a on the curve of f inverse.

   Here is why this reflection rule works, using the symmetry of the line y equals x.

   The line y equals x swaps x and y coordinates for any point reflected across it.

   Since f inverse of b equals a exactly when f of a equals b, swapping coordinates swaps the two curves.

   Now let's draw the inverse curve step by step in a worked example.

   Let f be defined on the interval from 0 to plus infinity by f of x equals square root of x.

   Step one: draw the line y equals x, our mirror for the reflection.

   Step two: pick points on the curve of f, like 1,1 and 4,2.

   Step three: swap each point's coordinates to get points on the inverse curve.

   Step four: join these new points smoothly to draw the curve of f inverse, which is x squared.

   Notice the two curves look like mirror images across the dashed line y equals x.

   This matches what we know: f inverse of x equals x squared, since f of x equals square root of x.

   The method is always the same: draw y equals x, plot a few points of f, reflect them, then join.

   This graphical method works even when you cannot find a formula for f inverse.

   Keep this picture in mind: two curves, symmetric across the line y equals x.

   Next, we will put everything together in one full bac-style problem.

   Across which line are the curves of f and f inverse always symmetric?

6. 6. Worked bac-style problem

   Let's finish with a full worked bac-style problem, using everything we learned.

   This problem brings together continuity, monotonicity, the bijection theorem, the derivative, and the graph.

   Let f be defined on the interval from 1 to plus infinity by f of x equals x squared minus 2x plus 3.

   Question one: show f is continuous and strictly increasing on this interval.

   f is a polynomial, so it is continuous on the whole interval from 1 to plus infinity.

   The derivative of f is two x minus two, which is positive for x greater than or equal to 1.

   So f is strictly increasing on the interval from 1 to plus infinity.

   Question two: show f realizes a bijection from the interval from 1 to plus infinity onto an interval we must find.

   We compute f of 1 equals 2, and the limit of f as x tends to plus infinity is plus infinity.

   Since f is continuous and strictly increasing, f of this interval is the interval from 2 to plus infinity.

   By the bijection theorem, f has an inverse function f inverse from the interval from 2 to plus infinity, onto the interval from 1 to plus infinity.

   Question three: compute the derivative of f inverse at the point b equals 3.

   First, find a such that f of a equals 3, so a squared minus 2a plus 3 equals 3.

   Since a must be in the interval from 1 to plus infinity, we keep a equals 2.

   Next, the derivative of f at 2 equals two times two minus two, which is two.

   By the formula, the derivative of f inverse at 3 equals one over the derivative of f at 2.

   Question four: sketch the curve of f inverse using the curve of f.

   We draw the line y equals x, then reflect points of f across it.

   Joining these reflected points gives us the curve of f inverse, symmetric to f across that line.

   This whole problem used every idea from this chapter, from continuity to the final graph.

   The method stays the same for any similar problem: check conditions, find the image interval, differentiate, then sketch.

   Keep these four steps in mind for any bac exam question on inverse functions.

   This completes our full chapter on inverse functions.

   In this problem, what is the value of a such that f of a equals 3?

   That completes our full chapter on inverse functions, from monotonic functions to sketching the inverse curve.

   You now have the bijection theorem, the derivative rule, and the reflection method ready for the bac.

   Keep practicing with these two exercises to make every step automatic.