Question

find ∆y and dy for the equation y=5x^2+6x+6 at x=2 when ∆x=0.001​

Answer 1

Step-by-step explanation:

dy=10x+6, at x=2 then dy=26,

Answer 2

∆y and dy for the equation y = 5x² + 6x + 6 are similar has the same value that is 0.026 at x =2 when ∆x=0.001​.

Given that,

We have to find ∆y and dy for the quadratic equation y = 5x² + 6x + 6 at x=2 when ∆x=0.001​.

We know that,

Let us take

y = f(x) = 5x² + 6x + 6

Δy = y₂ - y₁ = f(x₂) - f(x₁)

x₁ = x = 2

x₂ we have to find

Δx = x₂ - x₁

0.001 = x₂ - 2

x₂ = 0.001 + 2

x₂ = 2.001

Then delta y is

Δy = f(x₂) - f(x₁) = [ 5(x₂)² +6(x₂) + 6 ] - [ 5(x₁)² + 6x₁ + 6 ]

Δy = [5(2.001)² + 6(2.001 ) + 6 ] - [ 5(4) + 6(2) +6]

Δy = [20.020 + 12.006 + 6] - [ 20+12+6]

Δy = 38.026-38

Δy = 0.026

Then differentiation

$\frac{d}{dx}$f(x) = f'(x)

dy = f'(x)dx

dy = (10x+6)× 0.001

dy = (10×2+6)×0001

dy = 26 × 0.001

dy = 0.026

Therefore, ∆y and dy for the equation y = 5x² + 6x + 6 has the same value that is 0.026.

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