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Chapter 6: Problem 55

If \(f(x)=x^{3}+2 x^{2}+3 x+4\) and \(g(x)\) is the inverse of \(f(x)\), find \(g^{\prime}(4)\).

Short Answer

Expert verified
The derivative of the inverse function at 4, \(g'(4)\), is \(1/3\).

Step by step solution

01

Differentiate the function f(x)

The function given is \(f(x) = x^3 + 2x^2 + 3x + 4\). The derivative of this function can be found using the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\). So, the derivative of f(x) is \[f'(x) = 3x^{2} + 4x + 3\].
02

Find g(4) by setting f(x) equal to 4

The value of \(g(4)\) is equivalent to the \(x\) value in our original function at which \(f(x) = 4\). To find this, we solve the equation \[x^3 + 2x^{2} + 3x + 4 = 4\] for x. Simplifying, we get \[x^{3} + 2x^{2} + 3x = 0\]. Given that this is a cubic equation, it's hard to solve directly, but luckily it factors as \[x(x^2+2x+3)=0\], which means our solution is \(x = 0\). Hence, \(g(4)=0\).
03

Substitute into the inverse derivative formula

The derivative of the inverse function is given by the formula \(1/f'(g(x))\). Substituting our previously calculated \(f'(x)\) and \(g(4)\) into this formula, we get g'(4) as \[ g'(4) = 1/f'(g(4)) = 1/f'(0)\]. So \(g'(4)=1/(3(0)^{2} + 4*0 + 3)\] which gives \(g'(4)=1/3\).

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