Find each x-value at which the tangent line to f\left(x\right)=x^3-6x^2-34x+40 f ( x ) = x 3 − 6 x 2 − 34 x + 40 has slope 2.

  x ∈ {-2, 6}

Step-by-step explanation:

You want the solutions to ...

  f'(x) = 2

Taking the derivative, you get ...

  f'(x) = 3x^2 -12x -34 = 2

Subtract 2 and factor:

  3x^2 -12x -36 = 0 = 3(x^2 -4x -12) = 3(x -6)(x +2)

The values of x that satisfy this equation are the values of x that make the factors be zero: x = 6, x = -2.

The slope of a tangent to the curve will be 2 at x=-2 and at x=6.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

idk

step-by-step explanation: i just don't