Given points #(4, 70), (6, 69), (8, 72), (10, 81)# on the graph of a function #f(x)#, how do you find an approximate value for #f'(x)# ?

Answer 1

#f'(8) ~~ 3#

We can approximate #f'(8)# by evaluating the slope of the function between the two nearest given points #(6, 69)# and #(10, 81)#, namely:
#(81-69)/(10-6) = 12/4 = 3#

Bonus

Let's find a polynomial function that goes through these points and find out #f'(8)# for the resulting function.

Given points:

#(4, 70), (6, 69), (8, 72), (10, 81)#
Note that the #x# coordinates are evenly spaced, so let's look at the sequence of #y# values:
#color(blue)(70), 69, 72, 81#

Write down the sequence of differences between consecutive terms:

#color(blue)(-1), 3, 9#

Write down the sequence of differences between those differences:

#color(blue)(4), 6#

Write down the sequence of differences between those differences:

#color(blue)(2)#
Having arrived at a constant sequence (albeit of just one element), we can use the initial term of each of these sequences as a coefficient to give a formula for #f(x)#:
#f(x) = color(blue)(70)+(color(blue)(-1))/2(x-4) + (color(blue)(4))/(4*2)(x-4)(x-6)+(color(blue)(2))/(6*4*2)(x-4)(x-6)(x-8)#
#color(white)(f(x)) = 70-1/2x+2+1/2x^2-5x+12+1/24x^3-3/4x^2+13/3x-8#
#color(white)(f(x)) = 1/24 x^3 - 1/4 x^2 - 7/6 x + 76#

graph{1/24 x^3 - 1/4 x^2 - 7/6 x + 76 [-1, 12, 67, 83]}

Then:

#f'(x) = 1/8 x^2 - 1/2 x - 7/6#

and:

#f'(8) = 1/8 (8^2) - 1/2 (8) - 7/6 = 8-4-7/6 = 17/6#
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Answer 2

To find an approximate value for ( f'(x) ), the derivative of the function ( f(x) ), using the given points, you can use finite difference approximations. One commonly used method is the forward difference formula:

[ f'(x) \approx \frac{f(x + h) - f(x)}{h} ]

where ( h ) is the interval between the points. Since the points are evenly spaced with a difference of 2 between each x-value, we can use ( h = 2 ).

Using the forward difference formula with the given points:

  1. For the point (4, 70): [ f'(4) \approx \frac{f(4 + 2) - f(4)}{2} = \frac{f(6) - f(4)}{2} = \frac{69 - 70}{2} = -\frac{1}{2} ]

  2. For the point (6, 69): [ f'(6) \approx \frac{f(6 + 2) - f(6)}{2} = \frac{f(8) - f(6)}{2} = \frac{72 - 69}{2} = \frac{3}{2} ]

  3. For the point (8, 72): [ f'(8) \approx \frac{f(8 + 2) - f(8)}{2} = \frac{f(10) - f(8)}{2} = \frac{81 - 72}{2} = \frac{9}{2} ]

Now, you can use the same process to find ( f'(10) ). This will give you approximate values for the derivative ( f'(x) ) at each of the given points.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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