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acedstud
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Asked: 2021-03-14T10:19:24+00:00 In: ,

Question of the Day (14/03/2021)

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Find the equation of the tangent line to the curve f(x)=-3x-x^4 at x=-1.

derivative applications - tangent lineequation of tangent lineslope of tangent linetangent line

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2 Answers

  2. acedstud

    First, we find the slope of the tangent line by finding the derivative of f(x) at x=-1.

    Given: f(x)=-3x-x^4.

    Then, f'(x)=-3-4x^3 and f'(-1)=-3-4(-1)^3=-3-4(-1)=-3+4=1.

    Thus, the slope of the tangent line is 1.

    Next, we find the y-coordinate of the function f(x) at x=-1.

    At x=-1, y=f(-1)=-3(-1)-(-1)^4=3-1=2.

    Thus, the point of tangency is (-1, 2).

    Finally, the equation of a line with a slope of m=1 and passing through the point (x_1, y_1) = (-1, 2) is given as follows:

    y-y_1=m(x-x_1)

    \Rightarrow y-2=1(x-(-1))

    \Rightarrow y-2=x+1

    \Rightarrow y=x+1+2

    \Rightarrow y=x+3

    Therefore, the required equation is y=x+3.

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  4. qoraj
    Junior
    This answer was edited.

    We know that the equation of a tangent line of the curve on a function f at the point  A with abscissa x_0  is y=f'(x_0)(x-x_0)+f(x_0), in our case x_0=-1, so the equation is y=f'(-1)(x-(-1))+f(-1)

    y=f'(-1)(x+1)+f(-1)

    So let’s calculate  f'(-1) and f(-1).

    We have f(x)=-3x-x^4

    Thus f'(x)= -3 - 4x^3

    Then f'(-1)=-3 - 4(-1)^3 = -3-4(-1)=-3+4=1

    And f(-1)= -3(-1) - (-1)^4 =3 -1=2

    Therefore y=1(x+1)+2

    Finaly y=x + 3

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