Calculus Help!?

For which value of x on the graph of f(x) = (x+1)(x+2) such that the tangent is parallel to the line 3x - y - 5 = 0.

a) -1

b) 1

c) 3

d) 0

I'm confused on the question and how to figure out which answer is right. I'd appreciate if anyone can provide the correct answer and an understanding explanation. Thanks!

5 Answers

Relevance
  • 2 months ago

    3x-y-5=0

    =>

    y=3x-5 having the slope=3

    f(x)=(x+1)(x+2)

    =>

    f '(x)=(x+2)+(x+1)=2x+3

    f '(x)=3

    =>

    2x+3=3

    =>

    x=0

    The answer is that at x=0, the tangent

    of the curve is // to the given line.

  • 2 months ago

    f(x) = (x + 1)(x + 2) => x² + 3x + 2

    f '(x) = 2x + 3

    The line 3x - y - 5 = 0 can be written as:

    y = 3x - 5....hence, gradient 3

    So, we require where f '(x) = 3

    i.e. 2x + 3 = 3

    Hence, x = 0....i.e. at the point (0, 2)

    A sketch is below

    :)> 

    Attachment image
  • Philip
    Lv 6
    2 months ago

    f(x) = (x+1)(x+2) = x^2+3x+6. The tangent line to f(x) at P(x,f(x)) has slope f'(x);

    For what value of x is the tangent's slope equal to the slope of line L:3x-y-5 =0?

    I hope this explanation suffices to remove any confusion.;

    f'(x) = 2x + 3;

    Rewriting equation for L in slope, y-intercept form L: y = 3x-5 where 3 is the slope of L and the y-intercept of L is at (0,-5) gives us the slope of L.;

    We now put f'(x) = slope of L, ie., 2x+3 = 3, ie., x = 0.;

    So f'(0) = slope of L and the tangent line to f(x) is parallel to L when x = 0 and

    option d) gives the correct answer.

  • Pope
    Lv 7
    2 months ago

    I would expect you to learn properties of conics and quadratics before moving on to differential calculus, so let me suggest the following method.

    Any line parallel to 3x - y - 5 = 0 may be written in this form:

    y = 3x + k

    This equation gives the function graph:

    y = f(x)

    y = (x + 1)(x + 2)

    Substitute for y.

    3x + k = (x + 1)(x + 2)

    3x + k = x² + 3x + 2

    x² + 2 - k = 0

    That is a quadratic equation in x. Its solutions are the x-coordinates of any point lying on both the line and to function graph. The latter is a parabola, so there can be only one point lying on both. That means exactly one solution to the equation, so the discriminant is zero.

    (0)² - 4(1)(2 - k) = 0

    2 - k = 0

    k = 2

    Substitute that into the quadratic and solve for x.

    x² + 2 - k = 0

    x² + 2 - (2) = 0

    x² = 0

    x = 0

  • How do you think about the answers? You can sign in to vote the answer.
  • TomV
    Lv 7
    2 months ago

    The question reduces to, "What is the x value(s) of the point(s) where the derivative of f(x) = (x+1)(x+2) is equal to the slope of the line 3x-y - 5 = 0.

    Slope of the line:

    y = 3x-5 = mx+b

    m = 3

    Derivative of f(x) = (x+1)((x+2)

    f'(x) = x+1 + x+2 = 2x+3

    2x + 3  = 3

    2x = 0

    x = 0

    Ans: x = 0

Still have questions? Get your answers by asking now.