Answer :
To find the derivative of the function [tex]\( f(x) = -3x^2 - 6x \)[/tex] using limits, we can use the definition of the derivative.
The definition of the derivative of [tex]\( f(x) \)[/tex] at any point [tex]\( x \)[/tex] is given by:
[tex]\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \][/tex]
Let's start by calculating [tex]\( f(x+h) \)[/tex].
[tex]\[ f(x+h) = -3(x+h)^2 - 6(x+h) \][/tex]
First, expand [tex]\( (x+h)^2 \)[/tex]:
[tex]\[ (x+h)^2 = x^2 + 2xh + h^2 \][/tex]
Substitute this back into the function:
[tex]\[ f(x+h) = -3(x^2 + 2xh + h^2) - 6(x+h) \][/tex]
Distribute the terms:
[tex]\[ f(x+h) = -3x^2 - 6xh - 3h^2 - 6x - 6h \][/tex]
Now, we need to find the difference [tex]\( f(x+h) - f(x) \)[/tex]:
[tex]\[ f(x+h) - f(x) = (-3x^2 - 6xh - 3h^2 - 6x - 6h) - (-3x^2 - 6x) \][/tex]
Simplify the expression by canceling out all the terms that do not involve [tex]\( h \)[/tex]:
[tex]\[ f(x+h) - f(x) = -6xh - 3h^2 - 6h \][/tex]
Next, divide by [tex]\( h \)[/tex]:
[tex]\[ \frac{f(x+h) - f(x)}{h} = \frac{-6xh - 3h^2 - 6h}{h} = -6x - 3h - 6 \][/tex]
Finally, take the limit as [tex]\( h \)[/tex] approaches 0:
[tex]\[ f'(x) = \lim_{{h \to 0}} (-6x - 3h - 6) = -6x - 6 \][/tex]
So, the derivative function is:
[tex]\[ f'(x) = -6x - 6 \][/tex]
### Part b: Evaluate [tex]\( f'(a) \)[/tex] for the given values of [tex]\( a \)[/tex].
Evaluate the derivative at [tex]\( a = -2 \)[/tex]:
[tex]\[ f'(-2) = -6(-2) - 6 = 12 - 6 = 6 \][/tex]
Evaluate the derivative at [tex]\( a = 2 \)[/tex]:
[tex]\[ f'(2) = -6(2) - 6 = -12 - 6 = -18 \][/tex]
Therefore, the values of the derivative at the given points are:
[tex]\[ f'(-2) = 6 \][/tex]
[tex]\[ f'(2) = -18 \][/tex]
In summary:
- The derivative function [tex]\( f'(x) \)[/tex] is [tex]\( -6x - 6 \)[/tex].
- The evaluated derivatives are [tex]\( f'(-2) = 6 \)[/tex] and [tex]\( f'(2) = -18 \)[/tex].
The definition of the derivative of [tex]\( f(x) \)[/tex] at any point [tex]\( x \)[/tex] is given by:
[tex]\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \][/tex]
Let's start by calculating [tex]\( f(x+h) \)[/tex].
[tex]\[ f(x+h) = -3(x+h)^2 - 6(x+h) \][/tex]
First, expand [tex]\( (x+h)^2 \)[/tex]:
[tex]\[ (x+h)^2 = x^2 + 2xh + h^2 \][/tex]
Substitute this back into the function:
[tex]\[ f(x+h) = -3(x^2 + 2xh + h^2) - 6(x+h) \][/tex]
Distribute the terms:
[tex]\[ f(x+h) = -3x^2 - 6xh - 3h^2 - 6x - 6h \][/tex]
Now, we need to find the difference [tex]\( f(x+h) - f(x) \)[/tex]:
[tex]\[ f(x+h) - f(x) = (-3x^2 - 6xh - 3h^2 - 6x - 6h) - (-3x^2 - 6x) \][/tex]
Simplify the expression by canceling out all the terms that do not involve [tex]\( h \)[/tex]:
[tex]\[ f(x+h) - f(x) = -6xh - 3h^2 - 6h \][/tex]
Next, divide by [tex]\( h \)[/tex]:
[tex]\[ \frac{f(x+h) - f(x)}{h} = \frac{-6xh - 3h^2 - 6h}{h} = -6x - 3h - 6 \][/tex]
Finally, take the limit as [tex]\( h \)[/tex] approaches 0:
[tex]\[ f'(x) = \lim_{{h \to 0}} (-6x - 3h - 6) = -6x - 6 \][/tex]
So, the derivative function is:
[tex]\[ f'(x) = -6x - 6 \][/tex]
### Part b: Evaluate [tex]\( f'(a) \)[/tex] for the given values of [tex]\( a \)[/tex].
Evaluate the derivative at [tex]\( a = -2 \)[/tex]:
[tex]\[ f'(-2) = -6(-2) - 6 = 12 - 6 = 6 \][/tex]
Evaluate the derivative at [tex]\( a = 2 \)[/tex]:
[tex]\[ f'(2) = -6(2) - 6 = -12 - 6 = -18 \][/tex]
Therefore, the values of the derivative at the given points are:
[tex]\[ f'(-2) = 6 \][/tex]
[tex]\[ f'(2) = -18 \][/tex]
In summary:
- The derivative function [tex]\( f'(x) \)[/tex] is [tex]\( -6x - 6 \)[/tex].
- The evaluated derivatives are [tex]\( f'(-2) = 6 \)[/tex] and [tex]\( f'(2) = -18 \)[/tex].
