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\begin{cases} y=x^4 \\ y+8=a(x+5/4) \end{cases} How many solutions does this sytem have depending on parameter a? I need to solve it using derivatives somehow. Thank you.

I solved using hint which bubba gave me. Basically, we have this equation: $ x^4 + 8 = a(x + \frac54) $ Let's suppose that we have two functions: $ f(x) = x^4 + 8 $ and $g(x)=a(x+\frac54)$

$f(x)$ is a parabola looking thing while $g(x)$ is a line. So, we would have one solution when tangent line to $f(x)$ in some point $(x_0,f(x_0))$ derivative to $f(x)$ would be the same line as $g(x)$ So the derivative of $f(x)$ is $f'(x) = 4x^3$. So the tangent line would be $y = 4x_0^3(x-x_0) + f(x_0) = 4x_0^3x - 3x_0^4 + 8$. But we have that $y = g(x)$. So \begin{cases} a=4x_0^3 \\ \frac54a=8-3x_0^4 \end{cases}. Then we need to solve this equation: $5x_0^3=8-3x_0^4$ The solutions are $x_0 = -2$ and $x_0=1$ So, we would have one solution when $a = 4$ and $a=-32$

And we would have no solutions if $0<=a<4$ or $0>=a>-32$.

And we would have two solutions if $a>4$ or $a<-32$

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Yevs
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  • substracting this system gives you $-8 = x^4 - ax - \frac{5a}{4}$. Now you're down to one equation, can you take it from here? – DanielY Oct 13 '13 at 11:11
  • Welcome to MSE, Yevs. We ask that people show us what they've tried already. That said, here's a tiny hint: what can derivatives tell you about how a function behaves? – dfeuer Oct 13 '13 at 12:03
  • You can write up your solution as an answer (it's Ok to answer your own questions). You might get some up-votes, and gain some points. – bubba Oct 13 '13 at 23:56

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Think about the graphs of the two curves. The curve $y=x^4$ is a parabola-like thing. It's not really a parabola, but it has a similar shape. The other curve is a straight line through the point $(-\tfrac54, -8)$, whose slope depends on the value of $a$. For some values of $a$, the line will intersect the first curve twice, and for other values, there will be no intersection. The crucial state is when the two intersections coincide, which means that the line is tangent to the first curve. So, the key is to find a line that's tangent to the curve $y=x^4$ and passes through the point $(-\tfrac54, -8)$. As you said, you can use derivatives to do this.

bubba
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