Math Problem Statement

Let C be the graph of the function f(x) = (ax^2 + b) / x. Given that the line y = 2x - 1 is tangent to C at x = 1, find the values of a and b.

Solution

It appears that you have uploaded an image that includes a mathematical problem written in Arabic. I will extract the content to understand it better and provide a solution.

Let's translate the image and break down the question:

  1. The equation for the line is: y=2x1y = 2x - 1.
  2. The problem refers to the line being tangent to a curve CC, and it asks for conditions where the line CC is tangent at x=1x = 1.
  3. The curve CC is given by f(x)=ax2+bxf(x) = \frac{ax^2 + b}{x}.

Now, I will proceed with finding the conditions for the line to be tangent to the curve at x=1x = 1.

To do so:

  • The slope of the line at x=1x = 1 should be the same as the derivative of the curve at x=1x = 1.
  • The value of the curve at x=1x = 1 should be equal to the value of the line at x=1x = 1.

Let me calculate and solve for the parameters aa and bb. sympy import symbols, diff, Eq

Define variables

x, a, b = symbols('x a b')

Define the curve f(x) = (ax^2 + b) / x

f_x = (a * x**2 + b) / x

Calculate the derivative of f(x)

f_prime_x = diff(f_x, x)

The equation of the line is y = 2x - 1, and we want to find the conditions at x = 1

line_at_x_1 = 2 * 1 - 1 # Line value at x = 1 line_slope_at_x_1 = 2 # Slope of the line

The curve value at x = 1 should be equal to the line value at x = 1

curve_at_x_1 = f_x.subs(x, 1)

The slope of the curve at x = 1 should be equal to the slope of the line

curve_slope_at_x_1 = f_prime_x.subs(x, 1)

Now, solve the system of equations:

1. curve_at_x_1 = line_at_x_1

2. curve_slope_at_x_1 = line_slope_at_x_1

equation1 = Eq(curve_at_x_1, line_at_x_1) equation2 = Eq(curve_slope_at_x_1, line_slope_at_x_1)

Solve for a and b

from sympy import solve solution = solve([equation1, equation2], [a, b])

solutionThe values of aa and bb that make the line y=2x1y = 2x - 1 tangent to the curve f(x)=ax2+bxf(x) = \frac{ax^2 + b}{x} at x=1x = 1 are:

  • a=32a = \frac{3}{2}
  • b=12b = -\frac{1}{2}

Let me know if you need further details or clarification! Here are some follow-up questions you might find interesting:

  1. How can we generalize this approach for different tangent conditions?
  2. What is the geometric interpretation of a tangent line at a given point?
  3. How does the second derivative help in confirming the nature of tangency?
  4. Can you think of other types of functions that could have similar tangent conditions?
  5. What would happen if the curve were a polynomial instead of a rational function?

Tip: When solving for tangency, make sure to check both the value and the derivative at the point of interest!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Tangents
Algebra

Formulas

Derivative of a function: f'(x) = d/dx [f(x)]
Tangent condition: f(1) = 2(1) - 1
Slope matching condition: f'(1) = 2

Theorems

Derivative definition
Tangency condition

Suitable Grade Level

Grades 10-12