CALC question?? plz help!?
If g(x) = 1 – x^3, find g'(0).
Then use it to find an equation of the tangent line to the curve
y = 1 – x^3 at the point (0,1).
y = ___
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If g(x) = 1 – x^3
to find g'(0) you differentiate g(x) then sub x=0…
g'(x) = – 3x^2
sub x=0
g'(0) = -3(0)^2 = 0 ==>gradient of the tangent to the curve
to find the equation of the tangent, you use the formula y-y1 = m(x-x1) where (x1,y1)=(0,1)
this means y1 is the y coordinate ==>1
x1 is the x coordinate ==> 0
m is the gradient of the curve at (0,1) ==> 0
putting these into the formula you get
y – 1 = 0(x – 0)
y-1 = 0
therefore y = 1 is the final answer
0
Y=(1/2)-2x^3
0
solve g(x)
g'(x) = -3x^2
at x = 0, g’ = 0 …
the derivative of a function is its slope.
the slope is zero.
linear equation… y = mx + b m = 0….. y = b =1
y = 1
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