The line defined by the equation 2y+3=-(2/3)(x-3) is tangent to the graph of g(x) at x=-3. What is the value of the limit as x a
1 answer:
<span>the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>
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Equation of a tangent line of a curve is:
y- y0 = f´ (x0) (x - x0 ). In this case: x0=1, y0=1
f´(x)=
(Derivation)
f´(x0)=
y - 1 = 1/2 ( x - 1 )
y - 1 = 1/2 x - 1/2
y = 1/2 x - 1/2 + 1
y =
y- y0 = f´ (x0) (x - x0 ). In this case: x0=1, y0=1
f´(x)=
f´(x0)=
y - 1 = 1/2 ( x - 1 )
y - 1 = 1/2 x - 1/2
y = 1/2 x - 1/2 + 1
y =