Question Video: Determining Whether a Given Function Is a One-to-One Function | Nagwa Question Video: Determining Whether a Given Function Is a One-to-One Function | Nagwa

# Question Video: Determining Whether a Given Function Is a One-to-One Function Mathematics • Second Year of Secondary School

## Join Nagwa Classes

Is the function π(π₯) = 2π₯Β³ + 7π₯Β² + 5, where π₯ β β, a one-to-one function?

04:28

### Video Transcript

Is the function π of π₯ is equal to two π₯ cubed plus seven π₯ squared plus five, where π₯ is an element of the real numbers, a one-to-one function?

In this question, weβre given a function π of π₯ which is a cubic polynomial. And weβre also given the domain of this function. Itβs all real values of π₯. We need to use this information to determine if our function π of π₯ is a one-to-one function. To answer this question, letβs start by recalling what we mean by a one-to-one function. Itβs a function where each element of the range of the function corresponds to exactly one element of the domain of the function.

Another way of saying this is if we have π evaluated at π₯ sub one is equal to π evaluated at π₯ sub two for any values π₯ sub one and π₯ sub two in the domain of π, then we must have that π₯ sub one is equal to π₯ sub two. We want to determine if this property holds true for the function π of π₯ over the domain given to us in the question. And thereβs a few different ways we could go about this. For example, we could set π evaluated at π₯ sub one to be equal to π evaluated at π₯ sub two. And then we could either try and show that at π₯ sub one must be equal to π₯ sub two or try and find two values where this does not hold.

And although this method could work, it will be quite difficult because our function is a cubic polynomial. So instead, weβre going to do this by sketching a graph of π of π₯. So to sketch π of π₯, we start by noting itβs a cubic polynomial with positive leading coefficient. However, this is not enough information to fully sketch our curve. We could, for example, try and factor this polynomial to find its π₯-intercepts, but this is not necessary.

Instead, weβre going to find the critical points of our function by using differentiation. And we recall this is where its derivative is zero. Since π of π₯ is a polynomial, we can do this term-by-term by using the power rule for differentiation, which tells us for any real constants π and π the derivative of ππ₯ to the πth power with respect to π₯ is π times ππ₯ to the π minus one. We multiply it by the exponent of π₯ and reduce this exponent by one. We then apply this to each term separately. In our first term, the exponent of π₯ is three. We multiply by this exponent and reduce the exponent by one. We get six π₯ squared. In the second term, the exponent of π₯ is two. We multiply by this exponent of two and reduce the exponent by one. We get 14π₯ to the first power which is just 14π₯.

And we can differentiate the third and final term by using the power rule for differentiation. However, this is a constant. It doesnβt vary as π₯ varies. So its derivative is zero. This gives us that π prime of π₯ is equal to six π₯ squared plus 14π₯. We want to find the critical points. So we need to find the values of π₯ where this is zero. We need to solve zero is equal to six π₯ squared plus 14π₯. And we can do this by taking out a shared factor of two π₯ from both of the terms on the right-hand side of this equation. This then gives us that zero is equal to two π₯ multiplied by three π₯ plus seven.

Finally, for a product to be equal to zero, one of the factors must be equal to zero. So either π₯ is equal to zero or π₯ is equal to negative seven over three. And this is now enough information to sketch the general shape of the curve π¦ is equal to π of π₯. Itβs a cubic polynomial with positive leading term, and it has two unique critical points. The critical point on the left when π₯ is negative seven over three will be a local maxima and the critical point when π₯ is zero will be a local minima. And this is now enough to show that π of π₯ is not an injected function. We can do this directly from this diagram.

We can do this by choosing any value between π evaluated at zero and π evaluated at negative seven over three. For example, we can sketch the line π¦ is equal to π evaluated at zero onto this diagram. We can see that our line intersects the curve at two distinct points. And in fact, we know this has to be true. When π₯ is zero, the function has a local minima, and it goes up to a local maxima when π₯ is negative seven over three. However, itβs a cubic polynomial. So as π₯ approaches negative β, our functionβs outputs will approach negative β. So there must be two points of intersection between the line π¦ is equal to π evaluated at zero and the curve π¦ is equal to π of π₯.

And this is just an application of the horizontal line test. So letβs call the π₯-coordinate of the first point of intersection π₯ sub one and the π₯-coordinate of the second point of intersection π₯ sub two. Then, we can see that π evaluated at π₯ sub one is equal to π evaluated at π₯ sub two because their π¦-coordinates are equal on the diagram. However, π₯ sub one is not equal to π₯ sub two because theyβre two distinct points.

Therefore, we were able to answer that no, itβs not true that the function π of π₯ is equal to two π₯ cubed plus seven π₯ squared plus five for all real values of π₯ is a one-to-one function.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions