Math Problem Statement

Use the formula for instantaneous rate of​ change, approximating the limit by using smaller and smaller values of​ h, to find the instantaneous rate of change for the given function at the given value. f left parenthesis x right parenthesis equals 2 x Superscript x​; xequals2 Question content area bottom Part 1 The instantaneous rate of change for the function at xequals2 is

Solution

To find the instantaneous rate of change of the function f(x)=2xf(x) = 2^x at x=2x = 2, we use the limit definition of the derivative, which is:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

For f(x)=2xf(x) = 2^x at x=2x = 2, this becomes:

f(2)=limh022+h22hf'(2) = \lim_{h \to 0} \frac{2^{2 + h} - 2^2}{h}

Now, let's approximate this limit by using smaller and smaller values of hh. I'll calculate it for h=0.1,0.01,h = 0.1, 0.01, and 0.0010.001 to get progressively closer to the actual rate of change.

We know f(2)=22=4f(2) = 2^2 = 4.

Now let's compute f(2+h)f(2 + h) for different values of hh:

  1. For h=0.1h = 0.1: f(2.1)=22.14.287f(2.1) = 2^{2.1} \approx 4.287 4.28740.12.87\frac{4.287 - 4}{0.1} \approx 2.87

  2. For h=0.01h = 0.01: f(2.01)=22.014.028f(2.01) = 2^{2.01} \approx 4.028 4.02840.012.82\frac{4.028 - 4}{0.01} \approx 2.82

  3. For h=0.001h = 0.001: f(2.001)=22.0014.0028f(2.001) = 2^{2.001} \approx 4.0028 4.002840.0012.82\frac{4.0028 - 4}{0.001} \approx 2.82

From these approximations, we see that the instantaneous rate of change approaches 2.82.8 as hh becomes smaller. Hence, the instantaneous rate of change for f(x)=2xf(x) = 2^x at x=2x = 2 is approximately:

f(2)2.8f'(2) \approx 2.8

Would you like further details or have any questions?

Here are five related questions for you to consider:

  1. How does the derivative of exponential functions differ from polynomial functions?
  2. Can you explain why we use smaller and smaller values of hh for the approximation?
  3. What is the significance of the derivative in real-world applications?
  4. How would the derivative change if we used a different base, such as 3x3^x?
  5. What does the instantaneous rate of change represent geometrically?

Tip: The derivative gives you the slope of the tangent line at any given point on the function graph.

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Math Problem Analysis

Mathematical Concepts

Calculus
Instantaneous Rate of Change
Limits
Exponential Functions

Formulas

f'(x) = lim(h → 0) [(f(x + h) - f(x)) / h]
f(x) = 2^x

Theorems

Limit Definition of Derivative

Suitable Grade Level

Grades 11-12, Early College