Values of M so 2 curves don't intersect.

[trollcast]

Homework Statement

2 curves f(x) and g(x) don't intersect, find the range of values of m can be.

Homework Equations

$$f(x)=3x^2 - 2$$
$$g(x) = mx-5$$

The Attempt at a Solution

Could I work out the 2 values for m that mean g(x) is a tangent to f(x) at some point and then the range will be between them?

PS. I've already solved it via another approach but I was just wondering if this would work as it was my first initial idea on solving it.

 

[jedishrfu]

You're on the right track, the limiting values for m making the g(x) line a tangent to f(x) is correct.

have you plotted 3x^2 - 2 ? that would give you a hint of what the values of m are.

for g(x) you know one one point for certain: the y-intercept

 

[trollcast]

You're on the right track, the limiting values for m making the g(x) line a tangent to f(x) is correct.

have you plotted 3x^2 - 2 ? that would give you a hint of what the values of m are.

for g(x) you know one one point for certain: the y-intercept

I tried it again and I must have made a mistake when I tried this the first time as I definitely didn't get this.

For g(x) to be a tangent of f(x) their gradients must be the same:
$$f'(x)=6x \\
g(x)=6x(x)-5 \\
g(x)=6x^2-5$$

And they also must intersect each other:$$
f(x)=g(x)\\
3x^2-2=6x^2-5\\
3x^2-3=0\\
x^2-1=0\\
x^2=1\\
x=\pm1$$
Then just work out the values of m for the values of x:$$
m=f'(x)\\
m=6x\\
m=6(\pm1)\\
m=\pm6\\

-6<m<6$$

 

[jedishrfu]

and you can check this by plotting the two lines and the original f(x).

 

[HallsofIvy]

I'm not sure you should focus on the slope. The two graphs will intersect when 3x^2- 2= mx- 5 which is the same as 3x^2- mx+ 3= 0.

The "discriminant" of that is (-m)^2- 4(3)(3)= m^2- 36= 0. That is a parabola that is negative for x between -6 and 6. Since a quadratic equation has no real solutions if the discriminant is negative, the graphs do not intersect for m between -6 and 6.

 

[trollcast]

I'm not sure you should focus on the slope. The two graphs will intersect when 3x^2- 2= mx- 5 which is the same as 3x^2- mx+ 3= 0.

The "discriminant" of that is (-m)^2- 4(3)(3)= m^2- 36= 0. That is a parabola that is negative for x between -6 and 6. Since a quadratic equation has no real solutions if the discriminant is negative, the graphs do not intersect for m between -6 and 6.

Thanks,

That was the method I used in class to solve it after I tried working out the tangents but made a mistake copying something.

Is there any advantage to that method, ie. would the tangents not work for some similar problems?