For f(x)=3x4−12x3+5 find the following.(A) f′(x)(B) The slope of the graph of f at x=2(C) The equation of the tangent line at x=2(D) The value(s) of x where the tangent line is horizontal(A) f'(x) = ____(B) At x=2, the slope of the graph of f is ___(C) At x=2, the equation of the tangent line is y = ___(D) The tangent line is horizontal at x =_____(Use a comma to separate answers as needed.)

Question

For f(x)=3x4−12x3+5 find the following.(A) f′(x)(B) The slope of the graph of f at x=2(C) The equation of the tangent line at x=2(D) The value(s) of x where the tangent line is horizontal(A) f&#039;(x) = ____(B) At x=2, the slope of the graph of f is ___(C) At x=2, the equation of the tangent line is y = ___(D) The tangent line is horizontal at x =_____(Use a comma to separate answers as needed.)

Andrew Reyes · Accepted Answer

Step 1Given:f(x)=3x4−12x3+5Step 2A) To find f′(x)Differentiate the function with respect to xddxf(x)=ddx(3x4−12x3+5)f′(x)=12z3−36x2

Elaine Verrett · Accepted Answer

Step 3B) The slope of the f(x) at x=2Substitute x=2 in f′(x)f′(x)∣x=2=12(2)3−36(2)2=12(8)−36(4)=96−144=−48Step 4Therefore, the slope of f at x=2 is -48.

karton · Accepted Answer

Step 5 NSK C) The equation of the tangent line at x = 2 First find the y coordinates by substituting x=2 in the given function. f(2)=3(2)4−12(2)3+5 y=48-96+5 y=-43 Step 6 Therefore, the equation of the tangent line passing through (2,-43) with slope -48, is y-(-43)=-48(x-(-2)) y+43=-48x+96 y=-48x+96-43 y=-48x+53 Step 7 D) The value(s) of x where the tangent line is horizontal Set f′(x) equal to 0 and solve for x to find the x values at which the tangent line is horizontal to given function. 12x3−36x2=0 12x2(x−3)=0 12x2=0,(x−3)=0 x=0,(x−3)=0 x=0,x=3 Step 8 Therefore, the tangent line is horizontal at x=0,3

For f(x) = 3x^{4} - 12x^{3} +5 find the following. NS

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