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

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Is the function 𝑓(π‘₯) = 2π‘₯Β³ + 7π‘₯Β² + 5, where π‘₯ ∈ ℝ, a one-to-one function?

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Video Transcript

Is the function 𝑓 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, we’re given a function 𝑓 of π‘₯ which is a cubic polynomial. And we’re also given the domain of this function. It’s all real values of π‘₯. We need to use this information to determine if our function 𝑓 of π‘₯ is a one-to-one function. To answer this question, let’s start by recalling what we mean by a one-to-one function. It’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 𝑓 evaluated at π‘₯ sub one is equal to 𝑓 evaluated at π‘₯ sub two for any values π‘₯ sub one and π‘₯ sub two in the domain of 𝑓, 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 𝑓 of π‘₯ over the domain given to us in the question. And there’s a few different ways we could go about this. For example, we could set 𝑓 evaluated at π‘₯ sub one to be equal to 𝑓 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, we’re going to do this by sketching a graph of 𝑓 of π‘₯. So to sketch 𝑓 of π‘₯, we start by noting it’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, we’re going to find the critical points of our function by using differentiation. And we recall this is where its derivative is zero. Since 𝑓 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‘Ž and 𝑛 the derivative of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is 𝑛 times π‘Žπ‘₯ to the 𝑛 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 doesn’t vary as π‘₯ varies. So its derivative is zero. This gives us that 𝑓 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 𝑦 is equal to 𝑓 of π‘₯. It’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 𝑓 of π‘₯ is not an injected function. We can do this directly from this diagram.

We can do this by choosing any value between 𝑓 evaluated at zero and 𝑓 evaluated at negative seven over three. For example, we can sketch the line 𝑦 is equal to 𝑓 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, it’s a cubic polynomial. So as π‘₯ approaches negative ∞, our function’s outputs will approach negative ∞. So there must be two points of intersection between the line 𝑦 is equal to 𝑓 evaluated at zero and the curve 𝑦 is equal to 𝑓 of π‘₯.

And this is just an application of the horizontal line test. So let’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 𝑓 evaluated at π‘₯ sub one is equal to 𝑓 evaluated at π‘₯ sub two because their 𝑦-coordinates are equal on the diagram. However, π‘₯ sub one is not equal to π‘₯ sub two because they’re two distinct points.

Therefore, we were able to answer that no, it’s not true that the function 𝑓 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.

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