In a Taylor series expansion of e^x about x=2, the coefficient of (x-2)^4 is -; solve f'(x)=0 to obtain stationary point, find f''(x), and y=f(x) where f(x)=x^3-3x^2+24x+5.

Steps to Solve

1. Identify the Function to Analyze

The function given is ( f(x) = x^3 - 3x^2 + 24x + 5 ).

2. Find the First Derivative

To find stationary points, we first compute the derivative of the function: $$ f'(x) = 3x^2 - 6x + 24 $$

3. Set the First Derivative to Zero

Next, we set the first derivative equal to zero to find stationary points: $$ 3x^2 - 6x + 24 = 0 $$

4. Solve the Quadratic Equation

We can use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Where ( a = 3 ), ( b = -6 ), and ( c = 24 ).

Calculating the discriminant: $$ b^2 - 4ac = (-6)^2 - 4(3)(24) = 36 - 288 = -252 $$

Since the discriminant is negative, there are no real solutions, meaning there are no stationary points.

5. Check for Points of Inflection

To check for points of inflection, we need the second derivative: $$ f''(x) = 6x - 6 $$

Set the second derivative to zero to find points of inflection: $$ 6x - 6 = 0 $$

Solving gives: $$ x = 1 $$

6. Analyze the Second Derivative

Now we can evaluate ( f''(1) ): $$ f''(1) = 6(1) - 6 = 0 $$

However, to confirm it's a point of inflection, we check the sign of the second derivative around ( x = 1 ):

• For ( x < 1 ) (e.g., ( x = 0 )): $$ f''(0) = 6(0) - 6 = -6 < 0 $$

• For ( x > 1 ) (e.g., ( x = 2 )): $$ f''(2) = 6(2) - 6 = 6 > 0 $$

Since the sign changes, ( x = 1 ) is indeed a point of inflection.