MAT01 Mathematics for Management and Economics

a) Determine the points of the graph of f(x) where the slope of the tangent line is 2.

b) In which points of the graph of \(f(x)=x^3+3x^2-7.5x-4.5\) is the tangent parallel to the straight line g with the equation \(y = 1.5x\)?

Here, we solve part a). Part b) can be solved in a similar way.

Structure of the problem

In exercise 2, we had to determine the slope of the tangent to a graph in a given place x. Here, the problem is inverted: given the slope of a straight line, here = 2, find a place x where this straight line can be shifted so that it is the tangent of that graph in the point (x / f(x)).

There are situations where no such place exists. Just take g(x) = 5 as an example: in each point of its graph, the tangent must have a slope of g'(x) = 0, so a place where this is equal to 2 does not exist. Likewise, h(x) = 1.5x has a constant slope of 1.5, and since this is unequal 2, there is no solution to our problem.

Our function \(f(x)=2x^3-3x^2-10x+4\), however, looks different: since it it a 3rd degree polynomial, its slope is not constant, but varies itself with different values of x.

To solve our problem, we need to know a couple of things:

(i) What is a tangent? (The theory behind this is in block II, learning sequence 1, of the script)

(ii) What is the slope of a tangent? (See the same place in the script for the theoretical background.)

(iii) How do you determine the slope of a tangent of a function f(x), i.e. how do you find f'(x), for a given value of x ? (The theory behind this is in block II, learning sequence 2, of the script.)

(iv) How can the tangent of a function f(x) be represented graphically?
(There are exercises on this problem in block II, learning sequence 1, of the script. Or try this: derivative puzzle.)

Sketching the graph of the function

To solve our problem, it makes sense to have a look at the graph of the function \(f(x)=2x^3-3x^2-10x+4\).

The diagram shows that at x=-1 and at x=2, the slope of the graph is 2 (when measuring, don't forget that the axes are scaled differently!), and there doesn't seem to be another place on the x-axis where this is the case.

Drawing tangents is never precise, so we need calculus to be sure of our findings. What we know, however, is that there must be exactly two points of the graph where the slope is 2.
To practise drawing graphs, click here.

We now turn to differential calculus to find an exact solution. This does not make our graphical analysis obsolete. On the contrary: we keep our "graphical findings" in mind to compare them to the results of calculus. Yet calculus alone allows us to find a solution without any graphical tools.

Formal approach

To determine the points for which the tangent has a slope of 2, we proceed as follows:

(i) Make use of the first derivative f'(x) to find the matching place or the matching places on the x-axis.

(ii) Then, determine values f(x) at those places. This can be done through the equation \(f(x)=2x^3-3x^2-10x+4\).

Step (i):

The derivative function f'(x) shows us the slope of the tangent for any given value x. We need to know where this is exactly 2. This means that \(6x^2-6x-10\), which is the first derivative of f(x), must be 2. When setting this term equal to 2, we obtain the equation \(6x^2-6x-10=2\) (subtract 2 from both sides to see this).

This is a quadratic equation, and we have the means to solve it! Either put your pocket calculator to work (program QUADGL or QUADEQ) or make use of the formula for quadratic equations in block I, learning sequence 2, of the script. Both approaches lead to the solutions \( {x_1=2} \) and \( {x_2}=-1\)

This is only an intermediate result, however. We now know that there are two places on the x-axis where the slope of the tangent to the graph of f is 2. But we had to identify points on the graph of f, not just the x-coordinates of those points.

Step (ii):

This is the last part of our exercise, and it is the easiest one. Just replace x in the equation of f by 2, and then by -1. Here are the results:

\({y_1}=f(2)=2 \cdot 2^3-3 \cdot 2^2-10 \cdot 2+4=-12\)

\({y_2}=f(-1)=2 \cdot (-1)^3-3 \cdot (-1)^2-10 \cdot (-1)+4=9\)

This means that in the points \({P_1}( 2 / -12 )\) and \({P_2}( -1 / 9 )\) ), the tangents to the graph of f have a slope of 2. In any other point of the graph, the slopes of the tangents are different from 2.

Remark:

The following diagram shows once again the link between our function f (upper part) and its derivative function f' (lower part). As you can see, f'(x)=2 for both input values x=2 and x=-1 found in (i).

Exercise a)

\(f(x)=2x^3-3x^2-10 \cdot x+4\)

\(f'(x)=6x^2-6x-10=2\)

\(6x^2-6x-12=0\)

\( \Rightarrow \quad {x_1}=2 \quad {x_2}=-1\)

\({y_1}=f(2)=2 \cdot 2^3-3 \cdot 2^2-10 \cdot 2+4=-12\)

\({y_2}=f(-1)=2 \cdot (-1)^3-3 \cdot (-1)^2-10 \cdot (-1)+4=9\)

In the points \({P_1}( 2 / -12 )\) and \({P_2}( -1 / 9 )\), the tangents to the graph of f have a slope of 2.

Exercise b)

\(f(x)=x^3+3x^2-7.5x-4.5\)

\(f'(x)=3x^2+6x-7.5\) is the derivative function, i.e. the slope of the tangent

\(y(x) = 1.5x\)

\(y'(x) = 1.5\) is the slope of the given straight line.

slope of the tangent = slope of the straight line: \(3x^2+6x-7.5=1.5\)

\(3x^2+6x-9=0\)

\( \Rightarrow \quad {x_1}=1 \quad {x_2}=-3\)

\({y_1}=f(1)=-8\), \({y_2}=f(-3)=18\)

In the points \({P_1}(1/-8)\) and \({P_2}(-3/18)\) the tangent line is parallel to the graph of \(y = 1.5x\).