Answered

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If the function below is differentiable at x=3, find a+b

[tex]\left \{ {ax+b, x\ \textless \ 3} \atop {x^2-4x+8, x\geq 3}} \right.[/tex]

Sagot :

Answer:

a+b=1

Step-by-step explanation:

y = x² - 4x + 8   .... y = 3² - 4*(3) + 8 = 5

(3,5) is tangent point and function is differentiable at x=3

dy/dx 2x - 4 = 0

x=2  .. slope of tangent

ax + b must continuous to function x² - 4x + 8 at (3,5) and its slope=2

(y-5)/(x-3) = 2

y-5 = 2x - 6

y = 2x - 1 = ax +b

a = 2 and b = -1

a + b = 2 + -1 = 1

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