Video Transcript
Is the function 𝑓 of 𝑥 is equal to two 𝑥 cubed plus seven 𝑥 squared plus five, where 𝑥 is an element of the real numbers, a one-to-one function?
In this question, we’re given a function 𝑓 of 𝑥 which is a cubic polynomial. And we’re also given the domain of this function. It’s all real values of 𝑥. We need to use this information to determine if our function 𝑓 of 𝑥 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 𝑥 sub one is equal to 𝑓 evaluated at 𝑥 sub two for any values 𝑥 sub one and 𝑥 sub two in the domain of 𝑓, then we must have that 𝑥 sub one is equal to 𝑥 sub two. We want to determine if this property holds true for the function 𝑓 of 𝑥 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 𝑥 sub one to be equal to 𝑓 evaluated at 𝑥 sub two. And then we could either try and show that at 𝑥 sub one must be equal to 𝑥 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 𝑥. So to sketch 𝑓 of 𝑥, 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 𝑥-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 𝑥 is a polynomial, we can do this term-by-term by using the power rule for differentiation, which tells us for any real constants 𝑎 and 𝑛 the derivative of 𝑎𝑥 to the 𝑛th power with respect to 𝑥 is 𝑛 times 𝑎𝑥 to the 𝑛 minus one. We multiply it by the exponent of 𝑥 and reduce this exponent by one. We then apply this to each term separately. In our first term, the exponent of 𝑥 is three. We multiply by this exponent and reduce the exponent by one. We get six 𝑥 squared. In the second term, the exponent of 𝑥 is two. We multiply by this exponent of two and reduce the exponent by one. We get 14𝑥 to the first power which is just 14𝑥.
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 𝑥 varies. So its derivative is zero. This gives us that 𝑓 prime of 𝑥 is equal to six 𝑥 squared plus 14𝑥. We want to find the critical points. So we need to find the values of 𝑥 where this is zero. We need to solve zero is equal to six 𝑥 squared plus 14𝑥. And we can do this by taking out a shared factor of two 𝑥 from both of the terms on the right-hand side of this equation. This then gives us that zero is equal to two 𝑥 multiplied by three 𝑥 plus seven.
Finally, for a product to be equal to zero, one of the factors must be equal to zero. So either 𝑥 is equal to zero or 𝑥 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 𝑥. It’s a cubic polynomial with positive leading term, and it has two unique critical points. The critical point on the left when 𝑥 is negative seven over three will be a local maxima and the critical point when 𝑥 is zero will be a local minima. And this is now enough to show that 𝑓 of 𝑥 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 𝑥 is zero, the function has a local minima, and it goes up to a local maxima when 𝑥 is negative seven over three. However, it’s a cubic polynomial. So as 𝑥 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 𝑥.
And this is just an application of the horizontal line test. So let’s call the 𝑥-coordinate of the first point of intersection 𝑥 sub one and the 𝑥-coordinate of the second point of intersection 𝑥 sub two. Then, we can see that 𝑓 evaluated at 𝑥 sub one is equal to 𝑓 evaluated at 𝑥 sub two because their 𝑦-coordinates are equal on the diagram. However, 𝑥 sub one is not equal to 𝑥 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 𝑥 is equal to two 𝑥 cubed plus seven 𝑥 squared plus five for all real values of 𝑥 is a one-to-one function.
