Math Problem Statement

Find all x values where the function f(x) = 6x^3 + 5x - 2 has a tangent line with a gradient equal to 6. Let F(x) = f(g(x)), g(4) = 6, g'(4) = 8, f'(4) = 2, and f'(6) = 10. Find the value of F'(4). Find the equation of the tangent line at the given value: f(x) = 3 cos x + πx at x = 0.

Solution

Let's go through each question step by step.

Problem 4

Find all xx values where the function f(x)=6x3+5x2f(x) = 6x^3 + 5x - 2 has a tangent line with a gradient equal to 6.

  1. To find the values of xx where the tangent has a gradient of 6, we need to calculate f(x)f'(x) and set it equal to 6.

  2. First, find f(x)f'(x): f(x)=6x3+5x2f(x) = 6x^3 + 5x - 2 Differentiating: f(x)=18x2+5f'(x) = 18x^2 + 5

  3. Set f(x)=6f'(x) = 6: 18x2+5=618x^2 + 5 = 6 18x2=118x^2 = 1 x2=118x^2 = \frac{1}{18} x=±118=±132=±26x = \pm \frac{1}{\sqrt{18}} = \pm \frac{1}{3\sqrt{2}} = \pm \frac{\sqrt{2}}{6}

Thus, the values of xx where the tangent line has a gradient of 6 are: x=±26x = \pm \frac{\sqrt{2}}{6}

Problem 5

Given F(x)=f(g(x))F(x) = f(g(x)), g(4)=6g(4) = 6, g(4)=8g'(4) = 8, f(4)=2f'(4) = 2, and f(6)=10f'(6) = 10, find the value of F(4)F'(4).

  1. To find F(4)F'(4), use the chain rule: F(x)=f(g(x))g(x)F'(x) = f'(g(x)) \cdot g'(x)

  2. Substitute x=4x = 4: F(4)=f(g(4))g(4)F'(4) = f'(g(4)) \cdot g'(4)

  3. From the information given:

    • g(4)=6g(4) = 6
    • g(4)=8g'(4) = 8
    • f(6)=10f'(6) = 10 (since g(4)=6g(4) = 6)
  4. Substitute these values: F(4)=f(6)8=108=80F'(4) = f'(6) \cdot 8 = 10 \cdot 8 = 80

So, F(4)=80F'(4) = 80.

Problem 6(a)

Find the equation of the tangent line at the given value f(x)=3cosx+πxf(x) = 3 \cos x + \pi x at x=0x = 0.

  1. First, evaluate f(0)f(0): f(0)=3cos(0)+π0=3f(0) = 3 \cos(0) + \pi \cdot 0 = 3

  2. Next, find f(x)f'(x): f(x)=3cosx+πxf(x) = 3 \cos x + \pi x Differentiating: f(x)=3sinx+πf'(x) = -3 \sin x + \pi

  3. Now, find f(0)f'(0): f(0)=3sin(0)+π=πf'(0) = -3 \sin(0) + \pi = \pi

  4. The equation of the tangent line is: y=f(0)+f(0)(x0)y = f(0) + f'(0)(x - 0) y=3+πxy = 3 + \pi x

Thus, the equation of the tangent line at x=0x = 0 is: y=3+πxy = 3 + \pi x


Would you like further details or explanations on any of these solutions?

Here are some related questions to explore:

  1. How would we find the points of inflection for the function f(x)=6x3+5x2f(x) = 6x^3 + 5x - 2?
  2. What is the process for finding the second derivative of a composite function like F(x)=f(g(x))F(x) = f(g(x))?
  3. How would the tangent line equation change if we used a different point for evaluation?
  4. Can you find other values of xx for which f(x)=3cosx+πxf(x) = 3 \cos x + \pi x has a horizontal tangent?
  5. What would be the derivative of f(x)=3cosx+πxf(x) = 3 \cos x + \pi x if it included a higher degree term?

Tip: When dealing with composite functions, always apply the chain rule carefully, evaluating inner and outer derivatives accurately to avoid errors.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Trigonometry
Tangent Line

Formulas

Derivative of a polynomial function
Chain Rule: F'(x) = f'(g(x)) * g'(x)
Slope of tangent line at a point
Trigonometric derivative: d/dx [cos(x)] = -sin(x)

Theorems

Chain Rule
Differentiation Rules

Suitable Grade Level

Grades 11-12