MHB -8.2.59 Find eq for level surface

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To find the equation for the level surface of the function f(x,y,z) = x + e^(y+z) that passes through the point (1, ln(4), ln(9)), the value of the function at that point is calculated as f(1, ln(4), ln(9)) = 37. Therefore, the equation of the level surface is x + e^(y+z) = 37. This represents all points (x, y, z) where the function equals 37. The discussion emphasizes the importance of substituting the given point into the function to derive the level surface equation. The final equation succinctly captures the relationship defined by the function at the specified point.
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$\tiny{x4.01}$
$\textsf{Find an equation for the level surface of the function}$
$$f(x,y,z)=x+e^{y+z}$$
$\textsf{ that passes through the point}$
$$(1, \ln(4), \ln(9))$$
 
Last edited:
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Basically, you want:

$$f(x,y,z)=f\left(1,\ln(4),\ln(9)\right)$$

So, what do you get?
 
MarkFL said:
Basically, you want:

$$f(x,y,z)=f\left(1,\ln(4),\ln(9)\right)$$

So, what do you get?

plug in would be

$f(1,\ln(4),\ln(9))=1+e^{\ln(4)+\ln(9)}=1+4 \cdot 9 =37$
 
Last edited:
karush said:
plug in would be

$f(1,\ln(4),\ln(9))=1+e^{\ln(4)+\ln(9)}=1+4 \cdot 9 =37$

Yes, so the equation of the level surface would be:

$$x+e^{y+z}=37$$
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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