Math Problem Statement

The function y = f(x) has a tangent at x = 3 that forms a 45° angle with the x-axis and passes through x = -6. The function g(x) is defined as g(x) = -x * f(2x - 1). Find g'(2).

Solution

The problem in the image is as follows:


Problem: The function y=f(x)y = f(x) has a tangent at x=3x = 3 that forms a 4545^\circ angle with the xx-axis and passes through the point x=6x = -6.

Additionally, the function g(x)g(x) is defined as: g(x)=xf(2x1).g(x) = -x \cdot f(2x - 1).

You are tasked to find the derivative of g(x)g(x) at x=2x = 2, i.e., g(2)g'(2).


Solution:

Step 1: Analyze the tangent line

The slope of a line forming a 4545^\circ angle with the xx-axis is: m=tan(45)=1.m = \tan(45^\circ) = 1.

So, at x=3x = 3, the derivative of f(x)f(x), f(3)f'(3), is: f(3)=1.f'(3) = 1.

The tangent line passes through the point (6,0)(-6, 0). Using the point-slope form of a line equation, the tangent line can be expressed as: yf(3)=m(x3).y - f(3) = m(x - 3). Since m=1m = 1, this becomes: yf(3)=(x3).y - f(3) = (x - 3). At x=6x = -6, the tangent line crosses the xx-axis, so: 0f(3)=63    f(3)=9.0 - f(3) = -6 - 3 \implies f(3) = 9.

Thus: f(3)=9,f(3)=1.f(3) = 9, \quad f'(3) = 1.


Step 2: Differentiate g(x)g(x)

The function g(x)g(x) is given as: g(x)=xf(2x1).g(x) = -x \cdot f(2x - 1).

To find g(x)g'(x), use the product rule: g(x)=ddx[xf(2x1)].g'(x) = \frac{d}{dx} \big[-x \cdot f(2x - 1)\big]. Applying the product rule: g(x)=(1)f(2x1)+(x)ddx[f(2x1)].g'(x) = (-1) \cdot f(2x - 1) + (-x) \cdot \frac{d}{dx} \big[f(2x - 1)\big].

The derivative of f(2x1)f(2x - 1) involves the chain rule: ddx[f(2x1)]=f(2x1)ddx(2x1)=f(2x1)2.\frac{d}{dx} \big[f(2x - 1)\big] = f'(2x - 1) \cdot \frac{d}{dx}(2x - 1) = f'(2x - 1) \cdot 2.

Thus: g(x)=f(2x1)2xf(2x1).g'(x) = -f(2x - 1) - 2x \cdot f'(2x - 1).


Step 3: Evaluate g(2)g'(2)

Substitute x=2x = 2 into g(x)g'(x): g(2)=f(2(2)1)2(2)f(2(2)1).g'(2) = -f(2(2) - 1) - 2(2) \cdot f'(2(2) - 1).

Simplify: g(2)=f(41)4f(41).g'(2) = -f(4 - 1) - 4 \cdot f'(4 - 1). g(2)=f(3)4f(3).g'(2) = -f(3) - 4 \cdot f'(3).

From Step 1, we know: f(3)=9,f(3)=1.f(3) = 9, \quad f'(3) = 1.

Substitute these values: g(2)=941=94=13.g'(2) = -9 - 4 \cdot 1 = -9 - 4 = -13.


Final Answer: g(2)=13.g'(2) = -13.


Would you like me to elaborate on any part of the solution?

5 Related Questions:

  1. How do we compute derivatives using the chain rule in general cases?
  2. What is the product rule for differentiation, and when is it applied?
  3. How does the slope of a tangent line relate to the derivative of a function?
  4. How can we verify the slope of a line given two points?
  5. What is the importance of defining composite functions like f(2x1)f(2x - 1)?

Tip: Always ensure that the slope of a tangent is consistent with the angle provided, as small errors here can lead to incorrect solutions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Chain Rule
Product Rule
Tangent Line

Formulas

Product Rule: (uv)' = u'v + uv'
Chain Rule: f(g(x))' = f'(g(x)) * g'(x)
Slope of a line forming 45° with x-axis: m = tan(45°) = 1

Theorems

Derivative rules
Point-slope form of a line equation

Suitable Grade Level

Grades 11-12 or college level