The derivative as a limit; slope of a tangent

Answer Key - Grade 12 Math Worksheet - Challenge Me

Circle the best answer, or write your answer on the line.

1. 1.

   Ari rationalizes [f(9 + h) − f(9)]/h for f(x) = 6√x and lets h → 0. What is f′(9)?

   1
2. 2.

   Bo rationalizes [f(4 + h) − f(4)]/h for f(x) = 12√x and lets h → 0. What is f′(4)?

   3
3. 3.

   Esme rationalizes [f(36 + h) − f(36)]/h for f(x) = 3√x and lets h → 0. What is f′(36)?

   1/4
4. 4.

   By the limit, f(x) = x² − 4x has f′(a) = 2a − 4. Kira wants the x where the tangent slope is 6. What is that x?

   5
5. 5.

   By the limit, f(x) = x² + 6x has f′(a) = 2a + 6. Luz wants the x where the tangent slope is 0. What is that x?

   -3
6. 6.

   By the limit, f(x) = x² − 3x has f′(a) = 2a − 3. Milo wants the x where the tangent slope is 5. What is that x?

   4
7. 7.

   Pia sees the limit of [(4 + h)³ − 64]/h as h → 0. It is f′(a) for one function f and one point a. Which pair is it?

   • f(x) = x³ at x = 4
   • f(x) = x³ at x = 64
   • f(x) = (4 + x)³ at x = 4
   • f(x) = x³ − 64 at x = 0
8. 8.

   Ren sees the limit of [√(9 + h) − 3]/h as h → 0. It is f′(a) for one function f and one point a. Which pair is it?

   • f(x) = √x at x = 9
   • f(x) = √x at x = 3
   • f(x) = √x − 3 at x = 0
   • f(x) = √(9 + x) at x = 9
9. 9.

   Tala sees the limit of [(2 + h)⁵ − 32]/h as h → 0. It is f′(a) for one function f and one point a. Which pair is it?

   • f(x) = x⁵ at x = 2
   • f(x) = x⁵ − 32 at x = 0
   • f(x) = (2 + x)⁵ at x = 2
   • f(x) = x⁵ at x = 32
10. 10.

    Zee sees the limit of [1/(5 + h) − 1/5]/h as h → 0. It is f′(a) for one function f and one point a. Which pair is it?

    • f(x) = 1/x at x = 1/5
    • f(x) = 1/(5 + x) at x = 5
    • f(x) = 1/x − 1/5 at x = 0
    • f(x) = 1/x at x = 5