Questions: Using the definition of the derivative, find f'(x). Then find f'(1), f'(2), and f'(3) when the derivative exists. f(x)=-x^2+6x-9 To find the derivative, complete the limit as h approaches 0 for (f(x+h)-f(x))/h. lim h -> 0 Find f'(x) using the definition of the derivative. f'(x)= Find f'(1). Select the correct answer below and, if necessary, fill in the answer box to complete your choice. A. f'(1)= (Type an integer or a simplified fraction.) B. The derivative does not exist. Find f'(2). Select the correct answer below and, if necessary, fill in the answer box to complete your choice. A. f'(2)= (Type an integer or a simplified fraction.) B. The derivative does not exist. Find f'(3). Select the correct answer below and, if necessary, fill in the answer box to complete your choice. A. f'(3)= (Type an integer or a simplified fraction.)

Solution Steps

To find the derivative of the function \( f(x) = -x^2 + 6x - 9 \) using the definition of the derivative, we need to compute the limit as \( h \) approaches 0 of the difference quotient \(\frac{f(x+h)-f(x)}{h}\). This will give us the expression for \( f^{\prime}(x) \). Once we have \( f^{\prime}(x) \), we can evaluate it at specific points \( x = 1, 2, \) and \( 3 \) to find \( f^{\prime}(1) \), \( f^{\prime}(2) \), and \( f^{\prime}(3) \).

To find the derivative \( f^{\prime}(x) \) of the function \( f(x) = -x^2 + 6x - 9 \), we use the definition of the derivative: \[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substituting \( f(x) \) into the expression, we have: \[ f(x+h) = -(x+h)^2 + 6(x+h) - 9 \] \[ = -(x^2 + 2xh + h^2) + 6x + 6h - 9 \] \[ = -x^2 - 2xh - h^2 + 6x + 6h - 9 \] The difference quotient becomes: \[ \frac{f(x+h) - f(x)}{h} = \frac{(-x^2 - 2xh - h^2 + 6x + 6h - 9) - (-x^2 + 6x - 9)}{h} \] \[ = \frac{-2xh - h^2 + 6h}{h} \] \[ = -2x - h + 6 \] Taking the limit as \( h \to 0 \), we find: \[ f^{\prime}(x) = \lim_{h \to 0} (-2x - h + 6) = -2x + 6 \]

Now, we evaluate \( f^{\prime}(x) = -2x + 6 \) at \( x = 1, 2, \) and \( 3 \).

\[ f^{\prime}(1) = -2(1) + 6 = 4 \]

\[ f^{\prime}(2) = -2(2) + 6 = 2 \]

\[ f^{\prime}(3) = -2(3) + 6 = 0 \]

Final Answer