Question

Calculus please help me to do Number 2,3 and 4
DON'T DO NUMBER 1
if you can do it step by step.

Answer 1

2. The slope of the line tangent to $f(x)$ at $x=c$ is equal to $f'(c)$ (the value of the derivative at $c$). We have from (1a) that $f'(3)=5$ (I assume you've already shown this). Then from the point-slope formula, we get

$y-f(3)=f'(3)(x-3)\implies y-6=5(x-3)\implies\boxed{y=5x-9}$

3. This is just a matter of solving $f'(x)=2$:

$3x^2+8x+6=2\implies3x^2+8x+4=(3x+2)(x+2)=0\implies x=-\dfrac23,x=-2$

The corresponding points on the graph are $\boxed{\left(-\dfrac23,-\dfrac{203}{27}\right)}$ and $\boxed{(-2,-9)}$.

4. Take the definition of the derivative to be

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

Then for $f(x)=\sqrt{x+5}$ we have

$f(x+h)=\sqrt{x+h+5}$

so

$f'(x)=\displaystyle\lim_{h\to0}\frac{\sqrt{x+h+5}-\sqrt{x+5}}h=\lim_{h\to0}\frac{(x+h+5)-(x+5)}{h\left(\sqrt{x+h+5}+\sqrt{x+5}\right)}$

$f'(x)=\displaystyle\lim_{h\to0}\frac h{h\left(\sqrt{x+h+5}+\sqrt{x+5}\right)}$

$f'(x)=\displaystyle\lim_{h\to0}\frac1{\sqrt{x+h+5}+\sqrt{x+5}}$

$\boxed{f'(x)=\dfrac1{2\sqrt{x+5}}}$