Math Problem Statement

Which is true in the photo

Which is true in the photo?

Solution

Let's analyze and solve the problems displayed in the uploaded images:

Image 1:

The task is to find the equation of the tangent line to the function f(x)=3x25xf(x) = 3x^2 - 5x at the point where x0=2x_0 = -2.

Solution:

To find the equation of the tangent line to a function, we need two things:

  1. The slope of the tangent, which is the derivative of the function evaluated at x0=2x_0 = -2.
  2. The point through which the tangent passes, which is (x0,f(x0))(x_0, f(x_0)).

Step 1: Derivative of f(x)=3x25xf(x) = 3x^2 - 5x: f(x)=ddx(3x25x)=6x5f'(x) = \frac{d}{dx}(3x^2 - 5x) = 6x - 5 Now evaluate the derivative at x0=2x_0 = -2: f(2)=6(2)5=125=17f'(-2) = 6(-2) - 5 = -12 - 5 = -17 So, the slope of the tangent line is m=17m = -17.

Step 2: Find the point on the curve at x0=2x_0 = -2: f(2)=3(2)25(2)=3(4)+10=12+10=22f(-2) = 3(-2)^2 - 5(-2) = 3(4) + 10 = 12 + 10 = 22 Thus, the point on the curve is (2,22)(-2, 22).

Step 3: Equation of the tangent line: The equation of a line is given by: yy1=m(xx1)y - y_1 = m(x - x_1) Substitute m=17m = -17, x1=2x_1 = -2, and y1=22y_1 = 22: y22=17(x+2)y - 22 = -17(x + 2) Simplify: y22=17x34y - 22 = -17x - 34 y=17x12y = -17x - 12 Thus, the equation of the tangent line is y=17x12y = -17x - 12.

Correct answer: y=17x12y = -17x - 12.


Image 2:

The task is to find the critical points of the function f(x)=42x+5x2f(x) = 4 - 2x + 5x^2.

Solution:

Critical points occur where the derivative of the function is zero or undefined.

Step 1: Derivative of f(x)=42x+5x2f(x) = 4 - 2x + 5x^2: f(x)=ddx(42x+5x2)=2+10xf'(x) = \frac{d}{dx}(4 - 2x + 5x^2) = -2 + 10x

Step 2: Set the derivative equal to zero to find the critical points: 2+10x=0-2 + 10x = 0 Solve for xx: 10x=210x = 2 x=210=0.2x = \frac{2}{10} = 0.2

So, there is one critical point at x=0.2x = 0.2.

Step 3: Find the corresponding value of f(x)f(x): Since we are only asked for the xx-values of the critical points, we can stop here.

Correct answer: x=0.2x = 0.2, so the critical point is (0.2,0)(0.2, 0).


Let me know if you'd like further details or explanations!

5 Related Questions:

  1. How do you find the second derivative of a function, and what does it represent?
  2. What are inflection points, and how do you determine them?
  3. How do you find horizontal tangents of a function?
  4. What is the significance of the critical points in determining maxima and minima?
  5. How do you calculate the slope of a curve at a specific point?

Tip: Always simplify the derivative fully before solving for critical points or evaluating the slope!

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Math Problem Analysis

Mathematical Concepts

Derivative
Tangent Line
Quadratic Functions
Critical Points

Formulas

Derivative of a function: f'(x) = lim(h→0) [f(x+h) - f(x)]/h
Equation of a tangent line: y - y1 = m(x - x1)
Setting derivative to zero to find critical points: f'(x) = 0

Theorems

First Derivative Test for Critical Points
Slope of a tangent line

Suitable Grade Level

Grades 10-12