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

2. The slope of the line tangent to [tex]f(x)[/tex] at [tex]x=c[/tex] is equal to [tex]f'(c)[/tex] (the value of the derivative at [tex]c[/tex]). We have from (1a) that [tex]f'(3)=5[/tex] (I assume you've already shown this). Then from the point-slope formula, we get[tex]y-f(3)=f'(3)(x-3)\implies y-6=5(x-3)\implies\boxed{y=5x-9}[/tex]3. This is just a matter of solving [tex]f'(x)=2[/tex]:[tex]3x^2+8x+6=2\implies3x^2+8x+4=(3x+2)(x+2)=0\implies x=-\dfrac23,x=-2[/tex]The corresponding points on the graph are [tex]\boxed{\left(-\dfrac23,-\dfrac{203}{27}\right)}[/tex] and [tex]\boxed{(-2,-9)}[/tex].4. Take the definition of the derivative to be[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]Then for [tex]f(x)=\sqrt{x+5}[/tex] we have[tex]f(x+h)=\sqrt{x+h+5}[/tex]so[tex]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)}[/tex][tex]f'(x)=\displaystyle\lim_{h\to0}\frac h{h\left(\sqrt{x+h+5}+\sqrt{x+5}\right)}[/tex][tex]f'(x)=\displaystyle\lim_{h\to0}\frac1{\sqrt{x+h+5}+\sqrt{x+5}}[/tex][tex]\boxed{f'(x)=\dfrac1{2\sqrt{x+5}}}[/tex]