- #1
Sesse
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*opps fixing the latex code*
I have a test coming up so I tried solving some review questions but couldn't solve all of them. I will also post the ones which I am not sure whether they are correct or wrong so please spare some time to check them too.
1.
F(x)=<br /> \left{<br /> \begin{array}{cc}<br /> -x, & \mbox{ if } x<-1\\<br /> x^2, & \mbox{ if } -1 \leq x<1\\<br /> \frac{1}{x}, & \mbox{ if } x\geq 1<br /> \end{array}<br /> \right<br />
Use limit definition of the derivative to see if the fuction is differentiable at x=-1 and x=1
At x=-1
<br /> \lim_{x\rightarrow -1^-} \frac{f(x)-f(-1)}{x-(-1)}=-1
\lim_{x\rightarrow -1^+} \frac{f(x)+1}{x+1} = \frac{(x+1)(x-1)}{x+1}=x-1=-2
If this solution for x=-1 is correct then it is also correct for x=1 so I won't show my solution for that.
So the fuction is not differentiable at both points?
2. Probably correct.
3. y=(t^{2/3}).e^{sin(t)}+t
Find the slope at (0,0)
\frac{dy}{dx}=\left(\frac{2}{3}t^{-1/3}e^{sin(t)}+t^{2/3}e^{sin(t)}cos(t)\right)+1
\frac{dy}{dx}=1?
More will come. :)
I have a test coming up so I tried solving some review questions but couldn't solve all of them. I will also post the ones which I am not sure whether they are correct or wrong so please spare some time to check them too.
1.
F(x)=<br /> \left{<br /> \begin{array}{cc}<br /> -x, & \mbox{ if } x<-1\\<br /> x^2, & \mbox{ if } -1 \leq x<1\\<br /> \frac{1}{x}, & \mbox{ if } x\geq 1<br /> \end{array}<br /> \right<br />
Use limit definition of the derivative to see if the fuction is differentiable at x=-1 and x=1
At x=-1
<br /> \lim_{x\rightarrow -1^-} \frac{f(x)-f(-1)}{x-(-1)}=-1
\lim_{x\rightarrow -1^+} \frac{f(x)+1}{x+1} = \frac{(x+1)(x-1)}{x+1}=x-1=-2
If this solution for x=-1 is correct then it is also correct for x=1 so I won't show my solution for that.
So the fuction is not differentiable at both points?
2. Probably correct.
3. y=(t^{2/3}).e^{sin(t)}+t
Find the slope at (0,0)
\frac{dy}{dx}=\left(\frac{2}{3}t^{-1/3}e^{sin(t)}+t^{2/3}e^{sin(t)}cos(t)\right)+1
\frac{dy}{dx}=1?
More will come. :)
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