Questions: a) Find the instantaneous rate of change and equation of the tangent line to f(x)=1/x^2 at x=2. b) Find the instantaneous rate of change and equation of the tangent line to f(x)= -4x^2+3x at x=-1

a) Find the instantaneous rate of change and equation of the tangent line to f(x)=1/x^2 at x=2.
b) Find the instantaneous rate of change and equation of the tangent line to f(x)= -4x^2+3x at x=-1

Transcript text: a) Find the instantaneous rate of change and equation of the tangent line to $f(x)=\frac{1}{x^{2}}$ at $x=2$. b) Find the instantaneous rate of change and equation of the tangent line to $f(x)=$ $-4 x^{2}+3 x$ at $x=-1$

Solution

Solution Steps

Solution Approach

a) To find the instantaneous rate of change of $ f(x) = \frac{1}{x^2} $ at $ x = 2 $, we need to compute the derivative $ f'(x) $ and evaluate it at $ x = 2 $. The equation of the tangent line can then be found using the point-slope form of a line.

b) To find the instantaneous rate of change of $ f(x) = -4x^2 + 3x $ at $ x = -1 $, we need to compute the derivative $ f'(x) $ and evaluate it at $ x = -1 $. The equation of the tangent line can then be found using the point-slope form of a line.

Step 1: Instantaneous Rate of Change for $ f(x) = \frac{1}{x^2} $ at $ x = 2 $

To find the instantaneous rate of change, we compute the derivative: \[ f'(x) = -\frac{2}{x^3} \] Evaluating at $ x = 2 $: \[ f'(2) = -\frac{2}{2^3} = -\frac{2}{8} = -\frac{1}{4} \]

Step 2: Equation of the Tangent Line for $ f(x) = \frac{1}{x^2} $ at $ x = 2 $

Using the point-slope form of the line, the equation of the tangent line is: \[ y - f(2) = f'(2)(x - 2) \] Calculating $ f(2) $: \[ f(2) = \frac{1}{2^2} = \frac{1}{4} \] Thus, the equation becomes: \[ y - \frac{1}{4} = -\frac{1}{4}(x - 2) \] Simplifying, we find: \[ y = -\frac{1}{4}x + \frac{3}{4} \]

Step 3: Instantaneous Rate of Change for $ f(x) = -4x^2 + 3x $ at $ x = -1 $

The derivative is: \[ f'(x) = 3 - 8x \] Evaluating at $ x = -1 $: \[ f'(-1) = 3 - 8(-1) = 3 + 8 = 11 \]

Step 4: Equation of the Tangent Line for $ f(x) = -4x^2 + 3x $ at $ x = -1 $

Using the point-slope form again, we find: \[ y - f(-1) = f'(-1)(x + 1) \] Calculating $ f(-1) $: \[ f(-1) = -4(-1)^2 + 3(-1) = -4 - 3 = -7 \] Thus, the equation becomes: \[ y + 7 = 11(x + 1) \] Simplifying, we find: \[ y = 11x + 4 \]

Final Answer

The instantaneous rate of change and the equations of the tangent lines are:

For $ f(x) = \frac{1}{x^2} $ at $ x = 2 $:
- Instantaneous Rate of Change: $ -\frac{1}{4} $
- Tangent Line: $ y = -\frac{1}{4}x + \frac{3}{4} $
For $ f(x) = -4x^2 + 3x $ at $ x = -1 $:
- Instantaneous Rate of Change: $ 11 $
- Tangent Line: $ y = 11x + 4 $

Thus, the final boxed answers are: \[ \boxed{-\frac{1}{4}, \; y = -\frac{1}{4}x + \frac{3}{4}, \; 11, \; y = 11x + 4} \]

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