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Question Video: Finding the Graph of a Cubic Function by Finding Its Turning Points Using Differentiation Mathematics • 12th Grade

Use derivatives to identify which of the following is the graph of the function 𝑓(π‘₯) = βˆ’π‘₯Β³ + 6π‘₯Β² βˆ’ 9π‘₯ + 1. [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Use derivatives to identify which of the following is the graph of the function 𝑓 of π‘₯ equals negative π‘₯ cubed plus six π‘₯ squared minus nine π‘₯ plus one.

We have been given five possible graphs of the function and we could eliminate some of the options by recognizing that our function is a cubic and that the coefficient of π‘₯ cubed is negative. This means that as π‘₯ tends to negative ∞, the function will tend to positive ∞ and as π‘₯ tends to positive ∞, the function will tend to negative ∞. As a result, we can conclude that either option (D) or (E) will be correct. We could then find the points of intersection with the π‘₯-axis by setting 𝑓 of π‘₯ equal to zero and solving the cubic. Depending on how many real solutions there were, we could then determine whether graph (D) or (E) was correct.

Whilst this method would be perfectly valid, we are asked in this question to use derivatives. We know that we can use differentiation to find the critical points of a function and as a result determine which is the correct graph. The critical points for a function occur when its first derivative 𝑓 prime of π‘₯ is equal to zero or does not exist. Since our function 𝑓 of π‘₯ is a cubic polynomial, its derivative will be defined for all values of π‘₯. And we therefore don’t need to be concerned about possible points where the derivative may not exist.

We can obtain an expression for 𝑓 prime of π‘₯ using the power rule of differentiation. Differentiating term by term, we multiply by the original exponent and then reduce the exponent by one. This gives us 𝑓 prime of π‘₯ is equal to negative three π‘₯ squared plus 12π‘₯ minus nine, noting that differentiating any constant gives us zero.

Next, we need to set this expression for 𝑓 prime of π‘₯ equal to zero. Dividing through by negative three, this simplifies to π‘₯ squared minus four π‘₯ plus three equals zero. The quadratic on the left-hand side can be factored into two sets of parentheses. We have π‘₯ minus one multiplied by π‘₯ minus three. This is because negative one and negative three have a product of three and a sum of negative four. As the product of the factors equals zero, one or both of the factors must equal zero. We have π‘₯ equals one or π‘₯ equals three. This tells us that the π‘₯-coordinates of the critical values of our function are π‘₯ equals one and π‘₯ equals three. This is true of graphs (B), (C), and (D).

Whilst we established from the shape of the graphs that (B) and (C) could not be correct, we can confirm that graph (D) is the correct answer by substituting our values of π‘₯ into the function. This will give us the 𝑦-coordinates of the critical points. Firstly, we will calculate 𝑓 of one. This is equal to negative one cubed plus six multiplied by one squared minus nine multiplied by one plus one, which is equal to negative three.

Substituting π‘₯ equals three into 𝑓 of π‘₯, we have 𝑓 of three is equal to negative three cubed plus six multiplied by three squared minus nine multiplied by three plus one. This is equal to one. The critical points of the graph have coordinates one, negative three and three, one. This does indeed correspond to graph (D). And we can therefore conclude that this is the graph of the function 𝑓 of π‘₯ is equal to negative π‘₯ cubed plus six π‘₯ squared minus nine π‘₯ plus one.

Whilst it is not required in this question, we could go one stage further and confirm that the points one, negative three and three, one are local minimum and local maximum points, respectively. We can do this by considering the second derivative of our function. If the second derivative at that point is positive, we have a minimum. And if the second derivative is negative, we have a local maximum.

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