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Math  /  Calculus

QuestionThe table below gives selected values for the differentiable and increasing function ff and its derivative. If g(x)=f−1(x)g(x)=f^{-1}(x), what is the value of g′(−2)?g^{\prime}(-2) ? \begin{tabular}{|c|c|c|} \hlinexx & f(x)f(x) & f′(x)f^{\prime}(x) \\ \hline-2 & -5 & 4 \\ \hline 1 & -2 & 7 \\ \hline 4 & 3 & 3 \\ \hline 6 & 4 & 10 \\ \hline 7 & 6 & 9 \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. f f is a differentiable and increasing function.
2. g(x)=f−1(x) g(x) = f^{-1}(x) implies g g is the inverse function of f f .
3. We need to find g′(−2) g^{\prime}(-2) .

STEP 2

1. Use the property of inverse functions to find the relationship between derivatives.
2. Identify the necessary values from the table.
3. Calculate g′(−2) g^{\prime}(-2) .

STEP 3

Use the property of inverse functions:
For g(x)=f−1(x) g(x) = f^{-1}(x) , the derivative is given by:
g′(x)=1f′(g(x)) g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}

STEP 4

Identify g(−2) g(-2) :
Since g(x)=f−1(x) g(x) = f^{-1}(x) , we need f(g(−2))=−2 f(g(-2)) = -2 .
From the table, f(1)=−2 f(1) = -2 , so g(−2)=1 g(-2) = 1 .

STEP 5

Calculate g′(−2) g^{\prime}(-2) :
Using the formula from Step 1:
g′(−2)=1f′(g(−2))=1f′(1) g^{\prime}(-2) = \frac{1}{f^{\prime}(g(-2))} = \frac{1}{f^{\prime}(1)}
From the table, f′(1)=7 f^{\prime}(1) = 7 .
Thus, g′(−2)=17 g^{\prime}(-2) = \frac{1}{7} .
The value of g′(−2) g^{\prime}(-2) is:
17 \boxed{\frac{1}{7}}

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