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.
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!
- PinkgreenLv 72 months ago
y=3x-5 having the slope=3
The answer is that at x=0, the tangent
of the curve is // to the given line.
- Wayne DeguManLv 72 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
- PhilipLv 62 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.
- PopeLv 72 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
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- TomVLv 72 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