At birth, one puppy weighs 15/16 pounds. A second puppy weighs 1 and 1/8 pounds. How much more did the second puppy weigh?
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It looks like you're given
<em>F'(x)</em> = 3<em>x</em>² + 7
and
<em>F</em> (0) = 5
and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.
The first is done for you, <em>F</em> (0) = 5.
For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:
Then <em>F</em> (0.1) = 5.701, <em>F</em> (0.2) = 6.408, <em>F</em> (0.5) = 8.625, and <em>F</em> (2.0) = 27.
On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is
<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>
Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.
A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.
Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have
<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1
<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0) → <em>F</em> (0.1) ≈ 5.7
<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2
<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1) → <em>F</em> (0.2) ≈ 6.403
<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5
<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2) → <em>F</em> (0.5) ≈ 8.539
<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0
<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5) → <em>F</em> (2.0) ≈ 20.164