What are the inflections points of #y= e^(2x) - e^x #?

Answer 1

By the Chain Rule, the first derivative is #y'=2e^{2x}-e^{x}# and the second derivative is #y''=4e^{2x}-e^{x}#. Inflection points occur at the values of #x# where the second derivative changes sign (from positive to negative or negative to positive).

Setting #y''=0# leads to #4e^{2x}-e^{x}=0#. The left-hand side of this equation can be factored as #e^{x}(4e^{x}-1)=0#. Since #e^{x}# is never zero, it follows that we just need to solve #4e^{x}-1=0# to get #x=ln(1/4)=-ln(4)\approx -1.386#.

You can check that #y''=4e^{2x}-e^{x}# changes sign at this point by graphing it. Therefore, there is an inflection point at #x=-ln(4)#. The second coordinate of this point can be found by plugging it into the original function to get #e^{2\cdot ln(1/4)}-e^{ln(1/4)}=1/16-1/4=-3/16#. The inflection point is therefore #(-ln(4),-3/16)#.

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Answer 2

William Abdalla

To find the inflection points of $y = e^{2x} - e^x$ , we first find the second derivative and set it equal to zero to find the points of inflection.

First derivative: $\frac{dy}{dx} = 2e^{2x} - e^x$

Second derivative: $\frac{d^2y}{dx^2} = 4e^{2x} - e^x$

Setting the second derivative equal to zero: $4e^{2x} - e^x = 0$

Dividing both sides by $e^x$ : $4e^x - 1 = 0$

Adding 1 to both sides: $4e^x = 1$

Dividing both sides by 4: $e^x = \frac{1}{4}$

Taking the natural logarithm of both sides: $x = \ln\left(\frac{1}{4}\right)$

Solving for $x$ : $x = -\ln(4)$

Thus, the inflection point is $x = -\ln(4)$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.