Exponential and log functions (1 Viewer)

[tradewind]

i need help with these problems.How do you do these?

* Find the equation of the normal to the curve y = e^x at the point where x=3, in exact form

* Find the stationary point on the curve y=xe^x and determine its nature.
Find any points of inflexion and find values of y as x becomes very large or small. Hence sketch the curve

Any help is much appreciated

 

[CM_Tutor]

Originally posted by tradewind
* Find the equation of the normal to the curve y = e^x at the point where x=3, in exact form

y = e^x
dy/dx = e^x

When x = 3, m_tang = e³, and y = e³

So, the tangent is y - e³ = e³(x - 3)
y = e³x - 3e³ + e³

So, y = e³x - 2e³ in gradient-intercept form
or, e³x - y - 2e³ = 0 in general form

* Find the stationary point on the curve y=xe^x and determine its nature.
Find any points of inflexion and find values of y as x becomes very large or small. Hence sketch the curve

y = xe^x
dy/dx = e^x * 1 + x * e^x, using the product rule
= e^x(x + 1)

For a Stat. Pt., put dy/dx = 0: e^x(x + 1) = 0
The only solution of this equation is x = -1, as e^x > 0 for all x.
At x = -1, y = (-1)e^-1 = -1 / e.
So, the required stationary point is at (-1, -1 / e)

d²y/dx² = (x + 1) * e^x + e^x * (1 + 0) = e^x(x + 2)

At the stat. pt., d²y/dx² = e^-1(-1 + 2) = 1 / e > 0. Hence, (-1, -1 / e) is a MIN TP.

For inflexions, put d²y/dx² = 0: e^x(x + 2) = 0
The only solution of this equation is x = -2, as e^x > 0 for all x.
At x = -2, y = (-2)e^-2 = -2 / e².
So, the possible inflexion point is at (-2, -2 / e²)

Since e^x > 0 for all x, and (x + 2) is positive for x > -2 and negative for x < -2, it follows that
d²y/dx² > 0 for x > -2, and d²y/dx² < 0 for x < -2.
Since d²y/dx² changes sign around (-2, -2 / e²), it is a point of inflexion.

As x ---> + inf, y ---> + inf
As x ---> - inf, y ---> 0^-

With all this information, you should be able to sketch the curve, and if you can't, I'm sure that someone will post a picture of it.

 

[tradewind]

Thanks but

why is e^x > 0 for all x?

 

[CM_Tutor]

Originally posted by tradewind
why is e^x > 0 for all x?

Well, you could try lots of values on a calculator to convince yourself that this is true, or you could plot the graph.

Alternately, do you agree that e^x > 0 for x > 0? If you do, then consider eⁿ, where n < 0.
Now, let x = -n, and so x > 0. Also, e^x = e^-n = 1 / eⁿ.
It follows that e^x and eⁿ have the same sign, which must be (+), as we already knew e^x > 0 for x > 0.

Thus, since e⁰ = 1 > 0, it follows that e^x > 0 for all real x.

 

[tradewind]

thanks for the great help. it clicks now

 

[tradewind]

new questions

i've got a couple of questions that i don't understand how to do.

* (e^2x + 1)^7

* Find any stationary points on the curve y= x^2 e^2x and sketch the curve.

 

[CM_Tutor]

Originally posted by tradewind
i've got a couple of questions that i don't understand how to do.

* (e^2x + 1)^7

I'm assuming that this is meant to be: Find the derivative of (e^2x + 1)⁷.

In this case, the answer is d/dx[(e^2x + 1)⁷] = 7(e^2x + 1)⁶ * d/dx(e^2x + 1), using the chain rule
= 7(e^2x + 1)⁶ * 2e^2x
= 14e^2x(e^2x + 1)⁶

If the question was meant to be something else, then please restate the question.

* Find any stationary points on the curve y= x^2 e^2x and sketch the curve.

y = x²e^2x
dy/dx = e^2x * 2x + x² * 2e^2x, using the product rule
= 2xe^2x(x + 1)

For a Stat. Pt., put dy/dx = 0: 2xe^2x(x + 1) = 0
The only solutions of this equation are x = 0 and x = -1, as e^2x > 0 for all x.
At x = -1, y = (-1)²e^2(-1) = 1 / e².
At x = 0, y = (0)²e²⁽⁰⁾ = 0
So, the required stationary points are at (0, 0) and (-1, 1 / e²)

You can test the nature - you'll find that (0, 0) is a MIN TP and (-1, 1 / e²) is a MAX TP.
Also, As x ---> + inf, y ---> + inf
As x ---> - inf, y ---> 0⁺

With all this information, you should be able to sketch the curve.

 

[tradewind]

thanks so much, you have been a great help

 

[Collin]

tradewind, like CM_Tutor has advised it's useful to sketch a graph. Similarly this applies to many questions of Calculus nature where it's useful to sketch a graph or atleast know what the function looks like. For your concern of why is f(x) = e^x > 0 for all x, if a graph is sketched, it automatically becomes clear that for f; f(x) < 0 is not part of the range of the function.

Another way to confirm is to assume f(x) can < 0 for all x. Hence for example let e^x = -1 (i.e e^x can < 0). Trying to solve for x we obtain after utilising the definition of a logarithm: x = log (base) e (-1) which is undefined, hence a contradiction. Similarly, this applies to all values of f(x) < 0, so by proof by contradiction we conclude that f(x) can only > 0 for all x.