Questions: The first derivative of a continuous function y=f(x) is y'=x(x-18)^2. Find y'' and then use the graphing procedure to sketch the general shape of the graph of f. y''=3x^2-54x+324

Solution Steps

The first derivative of the function is given as \( y' = x(x-18)^2 \). We need to find the second derivative \( y'' \).

First, expand \( (x-18)^2 \): \[ (x-18)^2 = x^2 - 36x + 324 \]

Now, substitute back into the first derivative: \[ y' = x(x^2 - 36x + 324) = x^3 - 36x^2 + 324x \]

Differentiate \( y' \) to find \( y'' \): \[ y'' = \frac{d}{dx}(x^3 - 36x^2 + 324x) = 3x^2 - 72x + 324 \]

The problem states that the second derivative is \( y'' = 3x^2 - 54x + 324 \). However, our calculation shows \( y'' = 3x^2 - 72x + 324 \). There seems to be a discrepancy in the coefficient of the \( x \) term.

Final Answer