More Help Please! Derivatives!

[fdragon]

Ok suppose that the function f has a continuous second derivative for all x, and that f(0)= 2, f '(0)= -3, and f ''(0).

Let g be a function whose derivative is given by g'(x)= (e^(-2x))(3 f(x)+ 2f '(x)) for all x

A. Write an equation of th line tangent to the graph of f at the point where x= 0

B. Is there sufficient information to determine whether or not the graph of f has a point of inflection when x= 0? Explain your answer.

C. Given that g(0)= 4, write an equation of the line tangent to the graph of g at the point where x= 0.

D. Show that g''(x) = (e^-2x)(-6f(x) - f'(x) + 2f''(x)). Does g have a local maximum at x=0? Justify your answer.

Lol I really don't know where to begin......
I do know that for a... you need a point and the slope but I don't know how to find it.
Thanks![/code]

 

[soroban]

Re: Stuff With Derivatives!

Hello, fdragon!

You may kick yourself . . .

Given that \(\displaystyle f(x)\)has a continuous second derivative for all \(\displaystyle x\)
and that \(\displaystyle f(0)\,=\,2,\;f'(0)\,=\,-3,\;f''(0)\,=\,0.\)

(A) Write an equation of the tangent to the graph of \(\displaystyle f\) at the point where \(\displaystyle x\,=\,0\)

I do know that for (A) you need a point and the slope, but I don't know how to find it.

They gave them to you . . .

. . . \(\displaystyle \underbrace{f(0)\,=\,2}_{\text{point }(0,2)}\;\;\underbrace{f'(0)\,=\,-3}_{\text{slope: {m=-3}}}\;\;f''(0)\,=\,0.\)

 

[fdragon]

Lol... that's what I thought but I wasn't sure..... thanks.

So I just plugged it into an equation and got
y=-3(x-0)+2

Is that right?

 

[fdragon]

Any of the other problems? and if I got the tangent equation of the line right.

 

[soroban]

Hello again, fdragon!

Given that \(\displaystyle f\) has a continuous second derivative for all \(\displaystyle x\)
and that: \(\displaystyle f(0)\,=\,2,\;f '(0)\,=\,-3,\;f''(0)\,=\,0\)

Let \(\displaystyle g\) be a function whose derivative is given by:
. . \(\displaystyle g'(x)\:= \:e^{-2x}\left[3\cdot f(x)\,+\, 2\cdot f '(x)\right]\) for all \(\displaystyle x\)

A. Write an equation of th line tangent to the graph of \(\displaystyle f\) at the point where \(\displaystyle x\.=\.0\)

We have a point \(\displaystyle (0,2)\) and a slope \(\displaystyle m\,=\,-3\)

The equation is: \(\displaystyle \:y\,-\,2\:=\:-3(x\,-\,0)\;\;\Rightarrow\;\;y \:=\:-3x\,+\,2\)

B. Is there sufficient information to determine whether or not the graph of \(\displaystyle f\)
has a point of inflection when \(\displaystyle x\,=\,0\)? . Explain your answer.

I'll let someone else handle this one . . .

C. Given that \(\displaystyle g(0)\,=\,4\), write an equation of the line tangent to the graph of \(\displaystyle g\)
at the point where \(\displaystyle x\,=\,0\)

We have a point \(\displaystyle (0,4)\)

The slope is: \(\displaystyle \:g'(0) \:=\:e^0\left[3\cdot f(0)\,+\,2\cdot f'(0)\right]\:=\:1\cdot\left[3(2)\,+\,2(-3)\right] \:=\:0\)

The tangent is the horizontal line: \(\displaystyle \:y\:=\:4\)

D. Show that \(\displaystyle g''(x)\:=\:e^{-2x}\left[-6\cdot f(x)\,-\,f'(x)\,+\,2\cdot f''(x)\right]\)
Does \(\displaystyle g\) have a local maximum at \(\displaystyle x\,=\,0\)? . Justify your answer.

We have: \(\displaystyle \:g'(x) \:=\:e^{-2x}\left[3\cdot f(x)\,+\,2\cdot f'(x)\right]\)

Differentiate (product rule):
. . \(\displaystyle g''(x) \;=\;e^{-2x}\left[3\cdot f'(x)\,+\,2\cdot f''(x)\right]\,-\,2e^{-2x}\left[3\cdot f(x)\,+\,2\cdot f'(x)\right]\)

. . . . . . \(\displaystyle =\;e^{-2x}\left[3\cdot f'(x)\,+\,2\cdot f''(x)\,-\,6\cdot f(x)\,-\,4\cdot f'(x)\right]\)

. . . . . . \(\displaystyle =\;e^{-2x}\left[-6\cdot f(x)\,-\,f'(x)\,+\,2\cdot f''(x)\right]\)


We have: \(\displaystyle \:g''(0) \;=\;e^0\left[-6\cdot f(0)\,-\,f'(0)\,+\,2\cdot f''(0)\right]\)

. . . . . . . . . . . . \(\displaystyle =\;1\cdot\left[-6(2)\,-\,(-3)\,+\,2(0)\right]\;=\;-9\) (negative)

Hence, the graph of \(\displaystyle g(x)\) is concave down: \(\displaystyle \cap\)

. . Therefore, \(\displaystyle g(x)\) has a local maximum at \(\displaystyle x\,=\,0\).

 

[fdragon]

Whoa..... an amazing explanation.. thanks..... one question though lol

.... When differentiating g'(x)....
in the first and second line when you distributed -2e^-2x... where did the e^-2x go?