Math Problem Statement

The curve below represents a function f defined and differentiable over all real numbers. The points G(-2; 5) and H(0; 1) are on the graph of f, and the tangents at these points are horizontal.

Solution

Let's go through the exercises step by step:

Exercice 1:

The graph shows the function ff, defined and differentiable over the set of real numbers, with the following conditions:

  • Point G(2;5)G(-2; 5) and point H(0;1)H(0; 1) are on the curve.
  • The tangents at points GG and HH are horizontal.

Questions:

  1. Determine f(0)f(0) and f(2)f(-2), justifying your answers.

    • From the graph, at point H(0;1)H(0; 1), we have f(0)=1f(0) = 1.
    • At point G(2;5)G(-2; 5), we have f(2)=5f(-2) = 5.
  2. Determine f(0)f'(0) and f(2)f'(-2), justifying your answers.

    • The tangents at points GG and HH are horizontal, which means the derivative at these points is zero:
      • f(2)=0f'(-2) = 0
      • f(0)=0f'(0) = 0
  3. The function ff is defined by f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d. Determine aa, bb, cc, and dd.

    • (a) Find f(x)f'(x) in terms of a,b,c,da, b, c, d: f(x)=3ax2+2bx+cf'(x) = 3ax^2 + 2bx + c

    • (b) Determine cc and dd:

      • Using the point H(0;1)H(0; 1): f(0)=d=1f(0) = d = 1, so d=1d = 1.
      • Using the point G(2;5)G(-2; 5): Substitute into f(2)=8a4b2c+d=5f(-2) = 8a - 4b - 2c + d = 5. Given d=1d = 1, we get: 8a4b2c+1=5    8a4b2c=4    4a2bc=28a - 4b - 2c + 1 = 5 \implies 8a - 4b - 2c = 4 \implies 4a - 2b - c = 2
    • (c) Find two equations for aa and bb:

      • Using f(0)=0f'(0) = 0: From f(x)=3ax2+2bx+cf'(x) = 3ax^2 + 2bx + c, substitute x=0x = 0: c=0    c=0c = 0 \implies c = 0
      • From f(2)=0f'(-2) = 0: f(2)=12a4b+c=0f'(-2) = 12a - 4b + c = 0
      • Since c=0c = 0, the equation becomes: 12a4b=0    3a=b12a - 4b = 0 \implies 3a = b
    • (d) Determine aa and bb, and thus f(x)f(x):

      • Using 4a2b=24a - 2b = 2 and b=3ab = 3a: 4a2(3a)=2    4a6a=2    2a=2    a=14a - 2(3a) = 2 \implies 4a - 6a = 2 \implies -2a = 2 \implies a = -1 b=3a=3b = 3a = -3
      • Therefore, f(x)=x33x2+1f(x) = -x^3 - 3x^2 + 1.

Exercice 2:

Let f(x)=(x2+2x+1)exf(x) = (x^2 + 2x + 1)e^{-x}.

Questions:

  1. Show in two different ways that f(x)0f(x) \geq 0 for all real xx.

    • Rewrite f(x)=(x+1)2exf(x) = (x + 1)^2 e^{-x}. Since (x+1)20(x+1)^2 \geq 0 and ex>0e^{-x} > 0, the product is non-negative.
  2. Show that ff is differentiable over R\mathbb{R}.

    • The product and chain rule indicate ff is differentiable since both components are differentiable.
  3. Study the sign of ff' and deduce the variations of ff over R\mathbb{R}.

    • Given f(x)=(x2+1)exf'(x) = (-x^2 + 1)e^{-x}, solve f(x)=0f'(x) = 0 and analyze the intervals to determine increasing or decreasing behavior.
  4. Construct the graph of f(x)f(x).

Would you like more details on any specific part, or help with graph construction? Here are some further questions to explore:

  1. How does differentiability relate to the continuity of a function?
  2. How can you confirm the results of the derivative graphically?
  3. What are the implications of horizontal tangents for local maxima or minima?
  4. How does the behavior of the exponential function affect the function f(x)f(x)?
  5. How can the second derivative test be used to confirm concavity?

Tip: To sketch functions accurately, always find critical points, inflection points, and asymptotic behavior!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Cubic Functions
Polynomial Derivatives
Horizontal Tangents

Formulas

f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c

Theorems

Derivatives and Horizontal Tangents Theorem
Polynomial Function Behavior

Suitable Grade Level

Grades 11-12 (High School Calculus)