Question

1. (5pts) find any maxima or minima (function values and points x where they occur) of the following function $-2x^{2}+4x + 3$

Explanation:

Step1: Find the first derivative

The function is \( f(x) = -2x^2 + 4x + 3 \). The first derivative \( f'(x) \) is found using the power rule. For \( ax^n \), the derivative is \( nax^{n - 1} \). So, \( f'(x)=\frac{d}{dx}(-2x^2 + 4x + 3)=-4x + 4 \).

Step2: Find critical points

Set \( f'(x) = 0 \) to find critical points:
\( -4x + 4 = 0 \)
Subtract 4 from both sides: \( -4x=-4 \)
Divide both sides by -4: \( x = 1 \)

Step3: Find the second derivative

The second derivative \( f''(x) \) of \( f(x) = -2x^2 + 4x + 3 \) is \( f''(x)=\frac{d}{dx}(-4x + 4)=-4 \)

Step4: Determine if it's a maxima or minima

Since \( f''(x)=-4<0 \), the function has a local maximum at \( x = 1 \).

Step5: Find the function value at \( x = 1 \)

Substitute \( x = 1 \) into \( f(x) \): \( f(1)=-2(1)^2 + 4(1)+3=-2 + 4 + 3 = 5 \)

Answer:

The function \( f(x)=-2x^2 + 4x + 3 \) has a local maximum of \( 5 \) at \( x = 1 \). There is no local minimum.