Let Fx 6x2 4xa Use The Definition Of Derivative And The Proc ✓ Solved

Letfx 6x2 4xa Use The Definition Of Derivative And The Proced

Given the function f(x) = 6x² - 4x, this assignment requires us to use the definition of the derivative to find f'(x), then evaluate the instantaneous rate of change at a specific point, find the slope of the tangent line at that point, and identify the corresponding point on the graph.

Specifically, the tasks are:

1. Use the definition of the derivative to find f'(x).
2. Calculate the instantaneous rate of change of f(x) at x = -1.
3. Determine the slope of the tangent to the graph of y = f(x) at x = -1.
4. Find the coordinates of the point on the graph at x = -1.

Sample Paper For Above instruction

Step 1: Find the derivative using the definition

The definition of the derivative at a point x is:

f'(x) = lim_h→0 [(f(x+h) - f(x)) / h]

Given f(x) = 6x² - 4x, we compute f(x+h):

f(x+h) = 6(x+h)² - 4(x+h) = 6(x² + 2xh + h²) - 4x - 4h = 6x² + 12xh + 6h² - 4x - 4h

Now, f(x+h) - f(x) is:

(6x² + 12xh + 6h² - 4x - 4h) - (6x² - 4x) = 12xh + 6h² - 4h

Divide by h:

[(12xh + 6h² - 4h)] / h = 12x + 6h - 4

Taking the limit as h → 0:

f'(x) = lim_h→0 (12x + 6h - 4) = 12x - 4

Thus, the derivative is:

f'(x) = 12x - 4

Step 2: Find the instantaneous rate of change at x = -1

Substitute x = -1 into f'(x):

f'(-1) = 12(-1) - 4 = -12 - 4 = -16

Therefore, the instantaneous rate of change of f(x) at x = -1 is -16.

Step 3: Find the slope of the tangent line at x = -1

The slope of the tangent line is given by the derivative at that point:

slope = f'(-1) = -16

This means the tangent line at x = -1 has slope -16.

Step 4: Find the point on the graph at x = -1

Calculate y = f(-1):

f(-1) = 6(-1)² - 4(-1) = 6(1) + 4 = 6 + 4 = 10

The point on the graph at x = -1 is (-1, 10).

To summarize:

• The derivative is: f'(x) = 12x - 4
• Instantaneous rate of change at x = -1: -16
• Slope of tangent at x = -1: -16
• Point on the graph at x = -1: (-1, 10)

References

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