Second derivative test to find the relative maxima and relative minima

[Hypeboi604]

I'm not sure how to solve for f''(0) because i can't take the natural log of a negative number so i don't know how to continue with the question. so if i cant continue does that mean there's only a relative minimum?

 

[Hypeboi604]

Sorry i dont know if my post posted earlier but i dont think it did but anyways i need some help.
I'm not sure how to find f''(0) for this function f(x)=e^(x)+e^e^-(2x) because i cant take the natural log of a negative number??

 

[stapel]

I'm not sure how to find f''(0) for this function f(x)=e^(x)+e^e^-(2x) because i cant take the natural log of a negative number?

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Using the Second Derivative Test, find the relative maxima and relative minima, if any, of:

. . . . .\(\displaystyle f(x)\, =\, e^x\, +\, e^{-2x}\)

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My working:

. . .\(\displaystyle f'(x)\, =\, e^x\, \cdot\, 1\, +\, e^{-2x}\, \cdot\, (-2)\, =\, e^x\, -\, 2\, e^{-2x}\)

. . .\(\displaystyle 0\, =\, e^x\, -\, 2\, e^{-2x}\, \Rightarrow\, e^x\, =\, 2\, e^{-2x}\)

. . .\(\displaystyle \ln(e^x)\, =\, \ln(2\, e^{-2x})\, =\, \ln(2)\, +\, \ln(e^{-2x})\)

. . .\(\displaystyle x\, =\, \ln(2)\, +\, (-2x)\)

. . .\(\displaystyle 3x\, =\, \ln(2)\, \Rightarrow\, x\, =\, \dfrac{\ln(2)}{3}\)

Yes, but here's what might be a simpler way, depending upon one's personal preference:

. . . . .\(\displaystyle 0\, =\, e^x\, -\, 2\, e^{-2x}\)

. . . . .\(\displaystyle 2\, e^{-2x}\, =\, e^x\)

Now multiply through by e^(2x) which, by exponent rules, will cancel out the power on the left-hand side:

. . . . .\(\displaystyle 2\, e^{-2x + 2x}\, =\, e^{x + 2x}\)

. . . . .\(\displaystyle 2\, e^{0}\, =\, e^{3x}\)

. . . . .\(\displaystyle 2\, =\, e^{3x}\)

. . . . .\(\displaystyle \ln(2)\, =\, \ln(e^{3x})\, =\, 3x\)

...and so forth. But your way is fine, too!

Mathematically, what does this (one) critical point tell you? You know that e^(whatever) is always positive, so you've got no x-intercepts. You know that e^x is growing for x positive, and e^(-2x) is even bigger (going backwards) for x negative. So how many possible max/min points might you have? Do you expect them to be max's or min's?

. . .\(\displaystyle f''(x)\, =\, e^x\, \cdot\, 1\, -\, 2\, e^{-2x}\, \cdot\, -2\, =\, e^x\, +\, 4\, e^{-2x}\)

. . .\(\displaystyle 0\, =\, e^x\, +\, 4\, e^{-2x}\)

Well, we can try the multiplication thing again:

. . . . .\(\displaystyle 0\, =\, e^{x + 2x}\, +\, 4\, e^{-2x + 2x}\, =\, e^{3x}\, +\, 4\)

. . . . .\(\displaystyle -4\, =\, e^{3x}\)

So, yes, you're right; we're stuck with something we know doesn't solve in the real numbers. So what does that tell us, mathematically? Are there any potential inflection points? If not, then what does the sign of the second derivative tell us about concavity? Does this gibe with what we guessed from the first derivative (and from the quick graph that we sneaked on our calculators)?

 

[Hypeboi604]

so because we know that f''(x) can't be 0 we check f"(c) to see if it's a negative number or positive number. f"(ln2/3)=3.77.... because it's a positive number we know that the function f is concaving up and therefore (ln2/3) is the relative minimum and absolute minimum? sorry im bad at english ahaha, im not really sure what the question is even asking me to do ._.