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Chapter 1: Problem 31

If \(f(x)=1 / x^{5}\), compute \(f(-2)\) and \(f^{\prime}(-2)\).

Short Answer

Expert verified
The value of \(f(-2) = -\frac{1}{32}\) and the value of \(f'(-2) = -\frac{5}{64}\).

Step by step solution

01

- Evaluate the Function at Given Point

To find the value of the function at a specific point, substitute the given value into the function. Compute \(f(-2)\) by substituting \(x = -2\) into \(f(x) = \frac{1}{x^5}\):\[f(-2) = \frac{1}{(-2)^5} = \frac{1}{-32} = -\frac{1}{32}\]
02

- Compute the Derivative of the Function

First, find the derivative of the function \(f(x) = \frac{1}{x^5}\). Use the power rule for derivatives which states that \(\frac{d}{dx}[x^n] = nx^{n-1}\):\[f'(x) = \frac{d}{dx}[x^{-5}] = -5x^{-6} = -\frac{5}{x^6}\]
03

- Evaluate the Derivative at Given Point

Substitute the given value \(x = -2\) into the derivative function \(f'(x) = -\frac{5}{x^6}\) to find \(f'(-2)\):\[f'(-2) = -\frac{5}{(-2)^6} = -\frac{5}{64}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is the process of determining the output of a function for a given input. For a function like \(f(x) = \frac{1}{x^5} \), evaluating the function means substituting the given value into the function and simplifying the result.
Here's how you can evaluate the function at the point \( x = -2 \):
  • First, take the given function \( f(x) = \frac{1}{x^5} \).
  • Next, substitute \(-2\) for \(x\) in the function: \( f(-2) = \frac{1}{(-2)^5} \).
  • Calculate the power: \( (-2)^5 = -32 \).
  • Simplify: \( f(-2) = \frac{1}{-32} = -\frac{1}{32} \).
So, \( f(-2) = -\frac{1}{32} \).
Power Rule for Derivatives
The power rule is a basic and essential rule in calculus for finding the derivative of a function of the form \( x^n \). The rule states: \( \frac{d}{dx}[x^n] = nx^{n-1} \).
To apply this to the function \( f(x) = \frac{1}{x^5} \), first rewrite the function as a power of \(x\): \( f(x) = x^{-5} \).
Next, apply the power rule:
  • Identify the exponent: \(n = -5\).
  • Apply the power rule: \( \frac{d}{dx}[x^{-5}] = -5x^{-6} \).
  • Rewrite in fraction form: \( f'(x) = -\frac{5}{x^6} \).
So, the derivative of \( f(x) = \frac{1}{x^5} \) is \( f'(x) = -\frac{5}{x^6} \).
Evaluating Derivatives
Once you have the derivative of a function, you may need to evaluate it at a specific point. This gives the slope of the tangent line to the function at that point. To evaluate the derivative \( f'(x) = -\frac{5}{x^6} \) at \( x = -2 \), follow these steps:
  • Take the computed derivative: \( f'(x) = -\frac{5}{x^6} \).
  • Substitute \(-2\) for \(x\) in the derivative: \( f'(-2) = -\frac{5}{(-2)^6} \).
  • Calculate the power: \( (-2)^6 = 64 \).
  • Simplify: \( f'(-2) = -\frac{5}{64} \).
So, the value of the derivative at \( x = -2 \) is \( f'(-2) = -\frac{5}{64} \).
This tells us the rate at which the function \( f(x) \) is changing at \( x = -2 \).

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Most popular questions from this chapter

Estimate how much the function $$f(x)=\frac{1}{1+x^{2}}$$ will change if \(x\) decreases from 1 to \(.9 .\)

Determine whether each of the following functions is continuous and/or differentiable at \(x=1\). \(f(x)=\left\\{\begin{array}{ll}x^{3} & \text { for } 0 \leq x<1 \\ x & \text { for } 1 \leq x \leq 2\end{array}\right.\)

Find the indicated derivative. \(\frac{d y}{d x}\) if \(y=1\)

In Exercises 37-48, use limits to compute \(f^{\prime}(x)\). [Hint: In Exercises \(45-48\), use the rationalization trick of Example \(8 .]\) \(f(x)=3 x+1\)

Rates of Change Suppose that \(f(x)=4 x^{2}\). (a) What is the average rate of change of \(f(x)\) over each of the intervals 1 to 2,1 to \(1.5\), and 1 to \(1.1 ?\) (b) What is the (instantaneous) rate of change of \(f(x)\) when \(x=1 ?\)
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