Video: Using a Function’s Graph to Find Information about Its Roots

The given figure shows the graph of 𝑓(π‘₯) = π‘₯Β³ βˆ’ 4π‘₯Β² + 1. Use the graph to determine the number of solutions to the equation π‘₯Β³ = 4π‘₯Β² βˆ’ 1. Use the graph to determine the intervals in which the solutions to π‘₯Β³ = 4π‘₯Β² βˆ’ 1 lie.

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Video Transcript

The given figure shows the graph of 𝑓 of π‘₯ is equal to π‘₯ cubed minus four π‘₯ squared plus one. Use the graph to determine the number of solutions to the equation π‘₯ cubed equals four π‘₯ squared minus one. Use the graph to determine the intervals in which the solutions to π‘₯ cubed equals four π‘₯ squared minus one lie.

To answer this question, we’ll begin by considering the equation π‘₯ cubed equals four π‘₯ squared minus one. We’re looking to find the number of solutions to this equation. And if we were going to solve the equation, we would solve it a little like solving a quadratic. We’d get an expression in terms of π‘₯ that is then equal to zero.

To achieve this, we’ll begin by subtracting four π‘₯ squared from both sides of our equation. That gives us π‘₯ cubed minus four π‘₯ squared equals negative one. We’ll then add one to both sides. And we’ve now achieved an expression in terms of π‘₯ that is then equal to zero. So we have π‘₯ cubed minus four π‘₯ squared plus one. And that’s equal to zero.

So what’s the relationship between this equation and the equation for the graph given. Well, in the original equation for the graph, we have 𝑓 of π‘₯ equals π‘₯ cubed minus four π‘₯ squared plus one. But we’ve made π‘₯ cubed minus four π‘₯ squared plus one be equal to zero. So 𝑓 of π‘₯ has been replaced with zero. 𝑓 of π‘₯ is the output of the graph. And we often call it 𝑦. So where does 𝑦 equal zero on our graph?

The equation 𝑦 equals zero is also known as the π‘₯-axis. So to find the solutions for π‘₯ cubed equals four π‘₯ squared minus one, we need to find the points on the graph where 𝑦 or 𝑓 of π‘₯ equals zero. And that’s the points where the curve intersects the π‘₯-axis. We can see that this happens on one, two, and three separate occasions. So this means the equation 𝑓 of π‘₯ equals zero or π‘₯ cubed equals four π‘₯ squared minus one has three solutions.

The second part of this question asks us to determine the intervals in which the solutions lie. And we can make estimates to the solutions by reading the points where that graph intersects the π‘₯-axis. So let’s consider the scale on the π‘₯-axis. We see that five small squares represent five units. And if we divide through by five, we see that one square represents one unit. So we can add the values negative two, negative one, one, two, three, and four onto our graph as shown. And if we look carefully, we can see that our first solution lies roughly halfway between negative one and zero. So we say that π‘₯ is greater than negative one and less than zero.

It’s important here that we use a strict inequality because we can see that it definitely can’t be equal to negative one or zero. The next solution lies roughly halfway between zero and one. So we say that that solution is greater than zero and less than one. And our final solution lies between three and four. It’s closer to four than it is to three. But it doesn’t quite hit four. So we say that this solution π‘₯ is greater than three and less than four. So the intervals in which the solutions lie are π‘₯ is greater than negative one and less than zero, π‘₯ is greater than zero and less than one, and π‘₯ is greater than three and less than four.

Now, there is a way we can double check these intervals. Let’s take, for example, π‘₯ is greater than three and less than four. We can substitute π‘₯ equals three and π‘₯ equals four into the original function. And if we substitute three in, we get negative eight. And if we substitute one in, we get four.

Remember we said the solutions lie at the point where π‘₯ cubed minus four π‘₯ squared plus one is equal to zero. This sign change indicates to us that there must be a solution within this interval. And the graph is continuous, which is important because it means it doesn’t stop. We can guarantee that there will be a solution in the interval π‘₯ is greater than three and less than four. And we can repeat this process if we needed for π‘₯ is greater than negative one and less than zero and π‘₯ is greater than zero and less than one.

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