Example Question - specific point
Here are examples of questions we've helped users solve.
Finding the value of a function at a specific point
To find the value of f(5), we simply substitute x with 5 in the given function f(x) = -x^2 + 6x - 4. f(5) = -(5)^2 + 6(5) - 4 = -25 + 30 - 4 = 5 - 4 = 1 Thus, f(5) equals 1.
Calculating the Inverse of a Derivative at a Specific Point
To solve for the value of \((df/dx)^{-1}\) at the point \(x = 398 = f(5)\), we are essentially looking to compute the reciprocal of the derivative of \(f(x)\) at \(x = 5\). The reciprocal of the derivative is also known as the derivative of the inverse function evaluated at the corresponding y-value, which in this case is 398. First, we'll find \(df/dx\) by differentiating \(f(x)\): \(f(x) = 5x^3 - 9x^2 - 2\) \(df/dx = 15x^2 - 18x\) Now we need to evaluate \(df/dx\) at \(x = 5\): \(df/dx|_{x=5} = 15(5)^2 - 18(5) = 15(25) - 90 = 375 - 90 = 285\) Now, \((df/dx)^{-1}\) at \(x = 398\) is the reciprocal of this value: \((df/dx)^{-1}|_{f(5)} = 1 / df/dx|_{x=5} = 1/285\) So, the value is \(\frac{1}{285}\), which is the simplified fraction we were looking to find.
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