How do you find f'(2) using the limit definition given f(x)=9−x2?

f′(2)=limh→0f(2+h)−f(2)h=limh→0(9−(2+h)2)−(9−22)h=limh→0(9−(22+4h+h2))−(9−22)h=limh→0−4h+h2h=limh→0−4+h=−4 (Ans)

Using the limit definition, you get:f′(2)=limh→2f(x)−f(2)x−2=limh→2(9−x2)−(9−22)x−2=limh→29−x2−5x−2=limh→24−x2x−2=limh→2−x2−4x−2=limh→2−(x−2)(x+2)x−2=-(2+2)=-4

f′(2)=limh→0f(2+h)−f(2)h=limh→0(9−(2+h)2)−(9−22)h=limh→0(9−(22+4h+h2))−(9−22)h=limh→0−4h+h2h=limh→0−4+h=−4 (Ans)

Using the limit definition, you get:f′(2)=limh→2f(x)−f(2)x−2=limh→2(9−x2)−(9−22)x−2=limh→29−x2−5x−2=limh→24−x2x−2=limh→2−x2−4x−2=limh→2−(x−2)(x+2)x−2=-(2+2)=-4

Best solutions to - "How do you find f'(2) using the limit..."

High School
Calculus and Analysis
Calculus 1
Derivatives

Karlie Mays

Answered question

2022-04-16

How do you find f'(2) using the limit definition given

f (x) = 9 - x^{2}

Answer & Explanation

unduncjineei5r3

Beginner2022-04-17Added 19 answers

f^{'} (2) = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h}

= lim_{h \to 0} \frac{(9 - {(2 + h)}^{2}) - (9 - 2^{2})}{h}

= lim_{h \to 0} \frac{(9 - (2^{2} + 4 h + h^{2})) - (9 - 2^{2})}{h}

= lim_{h \to 0} \frac{- 4 h + h^{2}}{h}

= lim_{h \to 0} - 4 + h = - 4

(Ans)

prangijahnot

Beginner2022-04-18Added 17 answers

Using the limit definition, you get:

f^{'} (2) = lim_{h \to 2} \frac{f (x) - f (2)}{x - 2}

= lim_{h \to 2} \frac{(9 - x^{2}) - (9 - 2^{2})}{x - 2}

= lim_{h \to 2} \frac{9 - x^{2} - 5}{x - 2}

= lim_{h \to 2} \frac{4 - x^{2}}{x - 2}

= lim_{h \to 2} - \frac{x^{2} - 4}{x - 2}

= lim_{h \to 2} - \frac{(x - 2) (x + 2)}{x - 2}

=-(2+2)=-4

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