# Math Help - Calculus (Limits and Derivatives)

1. ## Calculus (Limits and Derivatives)

1. $lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}$

I reduced it to:

$lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}$

But I don't know what to do after that.

2. Find equations of all tangent lines to the graph of $y=4x^{3}+5x-8$

I took the derivative of the equation, which was:

$\acute{y}=12x^{2}+5$

I then thought I might need to make the derivative equal to the original equation, so:

$12x^{2}+5=4x^{3}+5x-8$

But I don't really know...

3. Determine values of constant real numbers a, b so that the function:

$f(x)= ax^{2}-8x+6$ if $x\leq-1$
$f(x)= bx+2$ if x is greater than -1

is differentiable at x=-1.

No idea how to do this one.

2. Originally Posted by Thomas
1. $lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}$

I reduced it to:

$lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}$

But I don't know what to do after that.
$\displaystyle\lim_{x\to-5}\frac{x^3+125}{(-5-x)^3}=-\lim_{x\to-5}\frac{(x+5)(x^2-10x+25)}{(5+x)^3}=$
$\displaystyle =-\lim_{x\to-5}\frac{x^2-10x+25}{(x+5)^2}=\frac{-100}{+0}=-\infty$

3. Ahh, I didn't think about how the bottom is squared.

Thanks!

4. Originally Posted by Thomas
3. Determine values of constant real numbers a, b so that the function:

$f(x)= ax^{2}-8x+6$ if $x\leq-1$
$f(x)= bx+2$ if x is greater than -1

is differentiable at x=-1.

No idea how to do this one.
First of all, f must be continuous at x=-1.
$\displaystyle\lim_{x\nearrow-1}f(x)=\lim_{x\nearrow-1}(ax^2-8x+6)=a+14=f(-1)$
$\displaystyle\lim_{x\searrow-1}f(x)=\lim_{x\searrow-1}(bx+2)=-b+2$.
Then $a+14=-b+2\Rightarrow a+b=-12$ (1)
We have $\displaystyle f'(x)=\left\{\begin{array}{cc}2ax-8, & x\leq-1\\b, & x>-1\end{array}\right.$

$f_l'(-1)=\lim_{x\nearrow-1}f'(x)=-2a-8, \ f_r'(-1)=\lim_{x\searrow-1}f'(x)=b$

$f_l'(-1)=f_r'(-1)\Rightarrow-2a-8=b\Rightarrow 2a+b=-8$ (2)

Now, solve the system formed by equations (1) and (2).