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Math  /  Calculus

Questiondydx(6y+2x)=4x−2y\frac{d y}{d x}(6 y+2 x)=4 x-2 y dydx=4x−2y6y+2x\frac{d y}{d x}=\frac{4 x-2 y}{6 y+2 x}
27. Let ff be the function defined by f(x)=x3+xf(x)=x^{3}+x. If g(x)=f−1(x)g(x)=f^{-1}(x) and g(2)=1g(2)=1, what is the value of g′(2)g^{\prime}(2) ? (D) 4 (E) 13 1=x3+x0=x2+x−1\begin{array}{l} 1=x^{3}+x \\ 0=x^{2}+x-1 \end{array}

Studdy Solution

STEP 1

1. We are given a function f(x)=x3+x f(x) = x^3 + x and its inverse g(x)=f−1(x) g(x) = f^{-1}(x) .
2. We know that g(2)=1 g(2) = 1 .
3. We need to find the derivative g′(2) g'(2) .

STEP 2

1. Verify the given condition g(2)=1 g(2) = 1 .
2. Use the relationship between a function and its inverse to find g′(x) g'(x) .
3. Evaluate g′(2) g'(2) .

STEP 3

Verify the given condition g(2)=1 g(2) = 1 .
Since g(x)=f−1(x) g(x) = f^{-1}(x) , it implies that if g(2)=1 g(2) = 1 , then f(1)=2 f(1) = 2 .
Calculate f(1) f(1) :
f(1)=13+1=1+1=2 f(1) = 1^3 + 1 = 1 + 1 = 2
The condition g(2)=1 g(2) = 1 is verified.

STEP 4

Use the relationship between a function and its inverse to find g′(x) g'(x) .
The derivative of the inverse function is given by:
g′(x)=1f′(g(x)) g'(x) = \frac{1}{f'(g(x))}
First, find f′(x) f'(x) :
f(x)=x3+x f(x) = x^3 + x f′(x)=3x2+1 f'(x) = 3x^2 + 1

STEP 5

Evaluate f′(g(2)) f'(g(2)) .
Since g(2)=1 g(2) = 1 , we need to find f′(1) f'(1) :
f′(1)=3(1)2+1=3×1+1=4 f'(1) = 3(1)^2 + 1 = 3 \times 1 + 1 = 4

STEP 6

Evaluate g′(2) g'(2) .
Using the formula for the derivative of the inverse:
g′(2)=1f′(g(2))=1f′(1)=14 g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)} = \frac{1}{4}
The value of g′(2) g'(2) is 14 \boxed{\frac{1}{4}} .

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