Solution :
Let y = f(x), x = 3 and x + δx. Then, δx.= 0.02.
For x = 3, we get
y = f(3) = 45
Now, y = f(x) ⟹ y = 3x2+5x+3
⟹ dydx = 6x + 5 ⟹ (dydx)x=3 = 23
Let δy be the change in y due to change δx in x. Then,
δy = dydx δx ⟹ δy = 23×0.02 = 0.46
f(3.02) = y + δy = 45 + 0.46 = 45.46
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