Use the Alternate form of the Limit Definition of Derivative to evaluate the derivative at the specified x-value.
a. f(x) = 4 – x ^ 2 at x = 3 f^ prime ( epsilon)+lim x+c (f(x) – f(epsilon))/(x – c) \ bm; x + 3 (- (x ^ 2 – a))/(x – 3) f’ * (3) = lim x -> 3 ((9x ^ 4) – f/2 * (3))/(x – 3) matrix 4\ 3 matrix (- (x + 3) * (x – 3))/(x – 3) matrix nm = x + 3 lim x+3 (9 – x ^ 2 – (4 – 3 ^ 3))/(x – 3) lim\ x + 3 – (x + 3) (9 – x ^ 2 – 4 + 9)/(x – 3) -(343) (9 – x ^ 2 – 4 + 9)/(x – 3) (- x ^ x + 9)/(x – 1) c. f(x) = |x + 5| x = – 5 (Hint: You will need to make use of one-sided limits.) f’ * c = lim x -> c (f(x) – f(c))/(x – c) lim x -5 ^ – |x+5| x+5 lim x -5^ + (1x + 51)/(x + 5) f^ prime (-5)=lim x -5 [|x + 5|] – f(- 5) x-(*5) matrix |-5^ * +5| -5^ * +5 |-5^ * -5| -5^ * +5 \ |- 5 + 5|/8 |-5^ * -5| 9^ * aligned lim v-5 [1×451]-0 x5 lim x -> – 5 |x + 5|/(x + 5) SPUT THE LIMIT -1 and 1… f is not differentiable at y -5
The Correct Answer and Explanation is :
Let’s break down the problem step by step and apply the limit definition of the derivative to each case. The derivative of a function ( f(x) ) at a specific point ( x = c ) is given by the limit definition of the derivative:
[
f'(c) = \lim_{x \to c} \frac{f(x) – f(c)}{x – c}
]
Problem (a) ( f(x) = 4 – x^2 ) at ( x = 3 )
The function is ( f(x) = 4 – x^2 ), and we are asked to find the derivative at ( x = 3 ).
First, calculate ( f(3) ):
[
f(3) = 4 – 3^2 = 4 – 9 = -5
]
Now, applying the limit definition:
[
f'(3) = \lim_{x \to 3} \frac{f(x) – f(3)}{x – 3} = \lim_{x \to 3} \frac{(4 – x^2) – (-5)}{x – 3}
]
[
f'(3) = \lim_{x \to 3} \frac{4 – x^2 + 5}{x – 3} = \lim_{x \to 3} \frac{9 – x^2}{x – 3}
]
Factor the numerator:
[
9 – x^2 = (3 – x)(3 + x)
]
Thus, the expression becomes:
[
f'(3) = \lim_{x \to 3} \frac{(3 – x)(3 + x)}{x – 3}
]
Notice that we can cancel out the ( (x – 3) ) term (since ( x \neq 3 )):
[
f'(3) = \lim_{x \to 3} -(3 + x)
]
Now substitute ( x = 3 ):
[
f'(3) = -(3 + 3) = -6
]
So, the derivative of ( f(x) ) at ( x = 3 ) is ( \boxed{-6} ).
Problem (b) ( f(x) = |x + 5| ) at ( x = -5 )
We are asked to find the derivative of ( f(x) = |x + 5| ) at ( x = -5 ). Since ( f(x) = |x + 5| ), the function is piecewise and behaves differently depending on whether ( x + 5 ) is positive or negative.
Case 1: ( x > -5 )
For ( x > -5 ), we have ( |x + 5| = x + 5 ). The derivative in this region is:
[
f'(x) = \frac{d}{dx}(x + 5) = 1
]
Case 2: ( x < -5 )
For ( x < -5 ), we have ( |x + 5| = -(x + 5) ). The derivative in this region is:
[
f'(x) = \frac{d}{dx}(-(x + 5)) = -1
]
At ( x = -5 ), the function is not differentiable.
To show this, we use the one-sided limits:
- Left-hand limit as ( x \to -5^- ):
[
\lim_{x \to -5^-} \frac{|x + 5| – |0|}{x + 5} = \lim_{x \to -5^-} \frac{-(x + 5)}{x + 5} = -1
] - Right-hand limit as ( x \to -5^+ ):
[
\lim_{x \to -5^+} \frac{|x + 5| – |0|}{x + 5} = \lim_{x \to -5^+} \frac{x + 5}{x + 5} = 1
]
Since the left-hand and right-hand limits are not equal, ( f'(x) ) does not exist at ( x = -5 ).
Thus, the derivative does not exist at ( x = -5 ), and we conclude that ( f(x) ) is not differentiable at ( x = -5 ).
Conclusion
- For ( f(x) = 4 – x^2 ) at ( x = 3 ), the derivative is ( f'(3) = -6 ).
- For ( f(x) = |x + 5| ) at ( x = -5 ), the derivative does not exist, since the left-hand and right-hand limits do not match.
This demonstrates the importance of considering one-sided limits when dealing with piecewise functions or absolute value functions.