# Tangent line and points it intersects a curve

1. Jan 6, 2012

### Kaldanis

This is another question from last year's calculus exam that I'm a bit stuck on.

" Let f be the real function given by f(x) = −x4+2x2+3x.
(a) Determine the tangent line to the curve y = f(x) at x = 1.
(b) Show that the tangent line to the curve y = f(x) at x = 1 is also the tangent line to another point on that curve and find the coordinates of that point.
(c) Find the subset of R on which f is convex. "​

My attempt:

(a) By putting x=1 into the original f(x)= −x4+2x2+3x, I get y=4. So I know the coordinates of this point are (1,4). I also know that f'(x)= -4x3+4x+3. By putting x=1 into this I get the gradient m=3. Putting all this into the line formula I get y-4=3(x-1), which rearranges to y=3x+1.

(b) Since I know the gradient of the line is 3, I just need to find the points where the derivative equals it. So I set -4x3+4x+3=3 and solve. Rearranging I get 4x(1-x2)=0. This only equals 0 when x= 0, 1 or -1. So I think this line is tangent to the points (0,0), (1,4) and (-1,-2). This is where I get confused, the question says there are two points on the curve where the line is tangent to it. I know (1,4) is definitely one, but I have two others to choose from?

(c) I'm pretty stuck on this. Am I right in thinking that if the second derivative is >=0, the function is convex upwards? If that is the case then I have to find when -12x²+4 >= 0? Even if that is the right way I'm having trouble working it out, I'm getting something like when -sqrt(4/12) < x < sqrt(4/12).

Please let me know if I'm along the right lines for 1 and 2 and direct me for 3? :)

Last edited: Jan 6, 2012
2. Jan 6, 2012

### eumyang

Yes, the slope at those points are the same (m = 3), but besides (1, 4), which point is on the tangent line y = 3x + 1? Part (b) states:

3. Jan 6, 2012

### Kaldanis

Ah okay, so once I know the different coordinates I just check to see which ones are on the tangent line. That means (1,4) and (-1,-2) are the answers. Makes sense, thanks!

4. Jan 6, 2012

### HallsofIvy

Staff Emeritus
Yes, the graph is convex when the second derivative is positive. That is, as you say, $-12x^2+ 4\ge 0$. That is the same as $-12x^2e\ge -4$ and then $x^2\le 1/3$ which is the same as what you give: $-1/\sqrt{3}\le x\le 1/\sqrt{3}$.

5. Jan 6, 2012

### Kaldanis

Thanks. This morning I had no idea how to do this question but I'm surprised at how straight forward it is. :)