take the derivative: = − 3 − 2x. Now, we can find the slope, m: = − 3 − 3(0) = − 3.Finally, we plug in the point (0, 4) and the slope m = −3 to get the equation of the tangent line:y − 4 = −3(x − 0) or y = − 3x + 4.- y = 0
Remember that the equation of a line through a point (x 1 , y 1 ) with slope m is y − y 1 = m (x −
x 1 ). We find the y-coordinate by plugging x = −2 into the equation y = (x^2 + 4x + 4)^2 , and we
find the slope by plugging x = −2 into the derivative of the equation.First, we find the y-coordinate, y 1 : y = ((−2)^2 + 4(−2) + 4)^2 = 0. This means that the linepasses through the point (−2, 0).Next, we take the derivative: = 2(x^2 + 4x + 4)(2x + 4). Now, we can find the slope, m: (2(−2) + 4) = 0. Finally, we plug in the point (−2, 0) and theslope m = 0 to get the equation of the tangent line: y − 0 = 0(x + 2) or y = 0.4.
The Mean Value Theorem says that if f(x) is continuous on the interval [a, b] and isdifferentiable everywhere on the interval (a, b), then there exists at least one number c on theinterval (a, b) such that . Here, the function is f(x) = x^3 + 12x^2 + 7x andthe interval is [−4, 4]. Thus, the Mean Value Theorem says that f′(x) = . This simplifies to f′(c) = 23. Next,
we need to find f′(c). The derivative of f(x) is f′(x) = 3x^2 + 24x + 7, so f′(c) = 3c^2 + 24c + 7.Now, we can solve for c: 3c^2 + 24c + 7 = 23 and . Note that is in the interval (−4, 4), but is not in the interval. Thus, the