This is another question from last year's calculus exam that I'm a bit stuck on.(adsbygoogle = window.adsbygoogle || []).push({});

" Let f be the real function given by f(x) = −x^{4}+2x^{2}+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)= −x^{4}+2x^{2}+3x, I get y=4. So I know the coordinates of this point are (1,4). I also know that f'(x)= -4x^{3}+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 -4x^{3}+4x+3=3 and solve. Rearranging I get 4x(1-x^{2})=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? :)

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# Tangent line and points it intersects a curve

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