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Find $\lambda\in\mathbb{R}$ such that $y=e^x$ and $y=\lambda x^2$ touch.

I'm just a beginner on derivatives, and I guess it should be done using them, but I'm totally stuck. (I guess it's not "touch" in English but oh well.)

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By touch do you mean they are tangent? Or that they intersect? –  Daniel Littlewood Jan 17 '13 at 20:23
    
They are tangent, that's the word. They always intersect anyway, on the left-hand side of $y$ axis, right? –  Lazar Ljubenović Jan 17 '13 at 20:25
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@LazarLjubenović Not if $\lambda = -1$.. –  Cocopuffs Jan 17 '13 at 20:26
    
Oh right it can be negative too... Gosh. –  Lazar Ljubenović Jan 17 '13 at 20:27
    
but if you set it negative, it is always disjoint from the exp, isnt it? I think it means tangent on the right side –  Zango Lotino Jan 17 '13 at 20:28

3 Answers 3

up vote 5 down vote accepted

Let me translate this problem:

You want to find a $\lambda \in \mathbb{R}$ so that there exists a point $a\in \mathbb{R}$ with the property:

$f(a)=g(a)$ (the curves intersect) and $f'(a)=g'(a)$ (their tangents are parallel match exactly) where $f(x)=e^x$ and $g(x)=\lambda x^2$. Can you do this?

Hint: $f'(x)=e^x=f(x)$ which means you want $g'(a)=g(a)$. Solve this for $\lambda,a$ and check it against the condition $f(a)=g(a)$

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Oh, I see - the slopes of tangents have to be the same, hence the derivatives have to be as well. Well that was easier than expected and I was just complicating it by myself. I assume the same logic would apply even if $f'(x)$ wasn't equal to $f(x)$, right? –  Lazar Ljubenović Jan 17 '13 at 20:32
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@LazarLjubenović Of course. The hint won't apply in that case however and things will be slightly more complicated –  Nameless Jan 17 '13 at 20:33

$y=e^{x}$ and $y=\lambda x^{2}$ are tangent at $x_{0}$ iff the curves and their derivatives are equal at $x_{0}$.

Thus, $e^{x_{0}}=\lambda x_{0}^{2}$ and $e^{x_{0}}=2\lambda x_{0}$. Therefore, $\lambda x_{0}^{2}=2\lambda x_{0} \implies x_{0}=2$. Substituting gives $\lambda=e^{2}/4$.
You can see this using graphing software.

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Notice that we could've also concluded that $x_{0}=0$, but this leads to an invalid result ($0=1$) –  Daniel Littlewood Jan 17 '13 at 20:40

Let $f(x)=e^x$ and $g(x)=\lambda x^2$. Then $f(x)$ and $g(x)$ are tangent if, at some point $x_0$:

$$f(x_0)=g(x_0)$$

(the curves meet) and:

$$f'(x_0)=g'(x_0)$$

(the curves have the same slope where they meet.)

What is $f'(x)$? What is $g'(x)$? This gives you two equations in two variables, $x_0$ and $\lambda$. Solve for $x_0$ first, and then solve for $\lambda$.

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