0

Consider $$\mathcal C: \quad g(x)=x^3-5x^2+1$$ and $\mathcal C$ its curve.

  1. Show that $\mathcal C$ has two tangents parallel to a line with equation $y=13x$.
  2. Find the points of tangency $E$ and $F$.

I can't think of a way to solve the first part but for part b I found the derivative of $g(x)$ and put it equal to $13$ and got $x=-1$ and $x=13/3$ (I don't know if it is correct though).

Sebastiano
  • 7,649
bluesky
  • 11
  • 1
    Welcome to Math SE. Please tell us what you have done to solve the problem so far. You will realise that questions like this with no work or attempt will usually be poorly received on this site. – kangzhe Feb 20 '21 at 17:05
  • So..... what do you want us to do? That's a standard derivative problem. Are you having trouble with it. Do you understand it? what concepts are you have haring trouble with? – fleablood Feb 20 '21 at 17:10
  • 1
    Being asked for "tangents parallel to a line with [a given equation]" tells you, via the slope of that line, a value of $c$ for which you are asked to solve $g'(x)=c$. The rest follows naturally out of that; you appear to have solved the problem. I admit that "find the points of tangency $E$ and $F$," is, without more context, potentially an ambiguous or confusing question. It is presumably asking for the ordered pairs $(a, g(a))$ and $(b, g(b))$ on the graph of $g$ which correspond to the two $x$-values $a$ and $b$ for which $g'(a) = g'(b) = c$, where $c$ is determined as in the first part.. – leslie townes Feb 20 '21 at 17:18

1 Answers1

1

  1. Your work for this part is correct so far. The derivative is $g'(x) = 3x^2 - 10x$. Recall that the equation $y = 13x$ has the slope-intercept form, which is a straight line with slope $13$. We solve for the value of $x$ such that $g'(x) = 13$. We have \begin{align} 3x^2 - 10x &= 13 \\ 3x^2 - 10x - 13 &= 0 \\ x &= -1 \text{ or } \frac{13}{3}. \end{align} This implies that at $x = -1$ and $x = \frac{13}{3}$ the tangents have a slope of $13$ which is parallel to $y = 13x$. If required by your course, you can check that these two values are on two different tangents indeed, though that will also answer part 2.

  2. To solve for the points of tangency $E$ and $F$ simply substitute the values of $x$ solved in part 1. into $g(x)$, that is,

$$ E = (-1, g(-1)) \quad F = \left(\frac{13}{3}, g\left(\frac{13}{3}\right)\right).$$

kangzhe
  • 133
  • 8