To find a function \( f(x) \) for which the average rate of change from \( x = -1 \) to \( x = 2 \) is equal to 5, we can use the formula for the average rate of change of a function:
\[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \]
Here, \( a = -1 \) and \( b = 2 \). So we have:
\[ \text{Average Rate of Change} = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{f(2) - f(-1)}{3} \]
We want this to equal 5:
\[ \frac{f(2) - f(-1)}{3} = 5 \]
Multiplying both sides by 3 gives:
\[ f(2) - f(-1) = 15 \]
Now, we need to find a function \( f(x) \) such that the difference \( f(2) - f(-1) = 15 \).
Example Function
Let's consider a linear function of the form:
\[ f(x) = mx + b \]
Compute \( f(2) \) and \( f(-1) \):
\[ f(2) = 2m + b \] \[ f(-1) = -m + b \]
Now, substituting these into the equation:
\[ f(2) - f(-1) = (2m + b) - (-m + b) = 2m + b + m - b = 3m \]
Setting this equal to 15:
\[ 3m = 15 \ m = 5 \]
So, we can have a function like:
\[ f(x) = 5x + b \]
The value of \( b \) can be any real number, so one possible function is:
\[ f(x) = 5x \]
Verification with \( f(x) = 5x \)
Let's verify:
\[ f(2) = 5(2) = 10 \] \[ f(-1) = 5(-1) = -5 \] \[ f(2) - f(-1) = 10 - (-5) = 10 + 5 = 15 \]
Thus, the average rate of change from \( x = -1 \) to \( x = 2 \) is indeed:
\[ \frac{15}{3} = 5 \]
Conclusion
One function that satisfies the condition is:
\[ f(x) = 5x \]
You can also have other functions that take the form \( f(x) = 5x + b \) for any constant \( b \).
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