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To find the maximum or minimum value of the function [tex]\( f(x) = 3 - 4x - x^2 \)[/tex], we need to follow these steps:
1. Find the derivative of [tex]\( f(x) \)[/tex]
The derivative of a function gives us the slope of the tangent line at any point on the function. Setting this derivative to zero helps us find the critical points, which are potential candidates for local maxima or minima.
[tex]\[ f'(x) = \frac{d}{dx}(3 - 4x - x^2) \][/tex]
Applying the rules of differentiation, we get:
[tex]\[ f'(x) = -4 - 2x \][/tex]
2. Set the derivative equal to zero to solve for the critical points
[tex]\[ -4 - 2x = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -2x = 4 \implies x = -2 \][/tex]
3. Determine the nature of the critical points using the second derivative test
The second derivative of a function helps determine whether a critical point is a maximum, minimum, or a point of inflection. The second derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f''(x) = \frac{d}{dx}(-4 - 2x) = -2 \][/tex]
The value of the second derivative is constant at [tex]\( -2 \)[/tex], which is negative. A negative second derivative indicates that the function is concave down at [tex]\( x = -2 \)[/tex], which means this critical point is a local maximum.
4. Evaluate the function at the critical point to find the maximum value
Substitute [tex]\( x = -2 \)[/tex] back into the original function:
[tex]\[ f(-2) = 3 - 4(-2) - (-2)^2 \][/tex]
Simplify:
[tex]\[ f(-2) = 3 + 8 - 4 = 7 \][/tex]
So, the function [tex]\( f(x) = 3 - 4x - x^2 \)[/tex] has a maximum value of 7 at [tex]\( x = -2 \)[/tex].
1. Find the derivative of [tex]\( f(x) \)[/tex]
The derivative of a function gives us the slope of the tangent line at any point on the function. Setting this derivative to zero helps us find the critical points, which are potential candidates for local maxima or minima.
[tex]\[ f'(x) = \frac{d}{dx}(3 - 4x - x^2) \][/tex]
Applying the rules of differentiation, we get:
[tex]\[ f'(x) = -4 - 2x \][/tex]
2. Set the derivative equal to zero to solve for the critical points
[tex]\[ -4 - 2x = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -2x = 4 \implies x = -2 \][/tex]
3. Determine the nature of the critical points using the second derivative test
The second derivative of a function helps determine whether a critical point is a maximum, minimum, or a point of inflection. The second derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f''(x) = \frac{d}{dx}(-4 - 2x) = -2 \][/tex]
The value of the second derivative is constant at [tex]\( -2 \)[/tex], which is negative. A negative second derivative indicates that the function is concave down at [tex]\( x = -2 \)[/tex], which means this critical point is a local maximum.
4. Evaluate the function at the critical point to find the maximum value
Substitute [tex]\( x = -2 \)[/tex] back into the original function:
[tex]\[ f(-2) = 3 - 4(-2) - (-2)^2 \][/tex]
Simplify:
[tex]\[ f(-2) = 3 + 8 - 4 = 7 \][/tex]
So, the function [tex]\( f(x) = 3 - 4x - x^2 \)[/tex] has a maximum value of 7 at [tex]\( x = -2 \)[/tex].
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