Hello, I have to do this exercise but I don't understand it. Could someone please help me?
a, b and c are three real numbers. Let f be the function defined in ℝ* by f(x)=ax+b+c/x .
Its representative curve is called C.
Determine the three real numbers a, b and c such that the curve C
-passes through A(2; 1)
-has a horizontal tangent at this point
-has at the point of abscissa 1 a tangent parallel to the line of equation y=(3/2)*x+2.
To check, the curve C is drawn here

Question

egr8066 egr8066

05-01-2024
Mathematics

Answered

Hello, I have to do this exercise but I don't understand it. Could someone please help me?
a, b and c are three real numbers. Let f be the function defined in ℝ* by f(x)=ax+b+c/x .
Its representative curve is called C.
Determine the three real numbers a, b and c such that the curve C
-passes through A(2; 1)
-has a horizontal tangent at this point
-has at the point of abscissa 1 a tangent parallel to the line of equation y=(3/2)*x+2.
To check, the curve C is drawn here

Answer :

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semsee45 semsee45 · Answer 1 · 2024-01-05T09:55:20-05:00

Answer:

[tex]a=-\dfrac{1}{2},\; b=3,\; c=-2[/tex]

Step-by-step explanation:

Given function:

[tex]f(x)=ax+b+\dfrac{c}{x}[/tex]

To find the values of the three real numbers a, b and c, we can use the given information to create a system of equations that we can then solve.

Equation 1

Given that the curve C of the function passes through point A(2, 1), we can plug in x = 2 and f(x) = 1 into the function to give:

[tex]\boxed{2a+b+\dfrac{c}{2}=1}[/tex]

Equation 2

The derivative of the function gives the slope of the tangent line at point (x, y).

Differentiate f(x):

[tex]f'(x)=a-\dfrac{c}{x^2}[/tex]

A horizontal tangent indicates that the derivative is zero at that point.

Given that the curve C has a horizontal tangent at point A(2, 1), we substitute x = 2 into f'(x) and set it equal to zero:

[tex]a-\dfrac{c}{2^2}=0[/tex]

[tex]\boxed{a-\dfrac{c}{4}=0}[/tex]

Equation 3

The point (1, y), where the x-coordinate is 1, has a tangent parallel to the line y = (3/2)x + 2, which has a slope of 3/2.

Given that the derivative of f(x) represents the slope of the tangent line at a point (x, y), we substitute x = 1 into f'(x) and set it equal to 3/2:

[tex]a-\dfrac{c}{1^2}=\dfrac{3}{2}[/tex]

[tex]\boxed{a-c=\dfrac{3}{2}}[/tex]

Now, we have a system of three equations with three unknowns:

[tex]\begin{cases}2a+b+\dfrac{c}{2}=1\\\\a-\dfrac{c}{4}=0\\\\a-c=\dfrac{3}{2}\end{cases}[/tex]

To find the values of a, b and c, solve the system of equations.

Rewrite equation 3 to isolate a:

[tex]a=\dfrac{3}{2}+c[/tex]

Substitute this into equation 2 and solve for c:

[tex]\begin{aligned}\dfrac{3}{2}+c-\dfrac{c}{4}&=0\\\\\dfrac{3}{4}c&=-\dfrac{3}{2}\\\\c&=-\dfrac{12}{6}\\\\c&=-2\end{aligned}[/tex]

Substitute the found value of c into equation 3 and solve for a:

[tex]\begin{aligned}a&=\dfrac{3}{2}-2\\\\a&=\dfrac{3}{2}-\dfrac{4}{2}\\\\a&=-\dfrac{1}{2}\end{aligned}[/tex]

Substitute the found values of a and c into equation 1 and solve for b:

[tex]\begin{aligned}2\left(-\dfrac{1}{2}\right)+b+\dfrac{(-2)}{2}&=1\\\\-1+b-1&=1\\\\b&=3\end{aligned}[/tex]

Therefore, the values of a, b and c are:

[tex]a=-\dfrac{1}{2},\; b=3,\; c=-2[/tex]