help with calculus question please
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BigGlenntheHeavy
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\(\displaystyle f(x) \ = \ x+x^{1/2}, \ f \ ' \ (x) \ = \ \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\displaystyle = \ \lim_{h\to0}\frac{(x+h)+(x+h)^{1/2}-x-x^{1/2}}{h}\)
\(\displaystyle = \ \lim_{h\to0}\frac{(h-x^{1/2})+(h+x)^{1/2}}{h}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{(h-x^{1/2})+(h+x)^{1/2}}{h} \ * \ \frac{(h-x^{1/2})-(h+x)^{1/2}}{(h-x^{1/2})-(h+x)^{1/2}}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{(h-x^{1/2})^{2}-h-x}{h[(h-x^{1/2})-(h+x)^{1/2}]}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{h^{2}-2hx^{1/2}+x-h-x}{h[(h-x^{1/2})-(h+x)^{1/2}]}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{h[h-2x^{1/2}-1]}{h[(h-x^{1/2})-(h+x)^{1/2}]}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{h-2x^{1/2}-1}{(h-x^{1/2})-(h+x)^{1/2}} \ = \ \frac{-2x^{1/2}-1}{-x^{1/2}-x^{1/2}} \ = \ \frac{2x^{1/2}+1}{2x^{1/2}}, \ QED\)
\(\displaystyle Now, \ domains \ and \ ranges.\)
\(\displaystyle f(x) \ = \ x+x^{1/2}, hence \ x \ \ge 0, \ therefore \ domain \ is \ [0,\infty) \ and \ range \ is \ [0,\infty).\)
\(\displaystyle f \ ' \ (x) \ = \ \frac{2 \sqrt x+1}{2 \sqrt x}, \ hence \ x \ > \ 0, \ so \ domain \ is \ (0,\infty) \ and \ range \ is \ (1,\infty)\)
\(\displaystyle Note: \ \lim_{x\to0^{+}}\frac{2 \sqrt x+1}{2\sqrt x} \ = \ \infty \ and \ \lim_{x\to\infty}\frac{2\sqrt x+1}{2\sqrt x} \ = \ 1.\)
\(\displaystyle See \ the \ graph \ below. \ Although \ not \ shown, \ x=0 \ is \ a \ vertical \ aysmptote \ and \ y=1 \ is \ a \ horizontal\)
\(\displaystyle \ aysmptote.\)
[attachment=0:mqqbw5pp]zzz.jpg[/attachment:mqqbw5pp]
\(\displaystyle = \ \lim_{h\to0}\frac{(x+h)+(x+h)^{1/2}-x-x^{1/2}}{h}\)
\(\displaystyle = \ \lim_{h\to0}\frac{(h-x^{1/2})+(h+x)^{1/2}}{h}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{(h-x^{1/2})+(h+x)^{1/2}}{h} \ * \ \frac{(h-x^{1/2})-(h+x)^{1/2}}{(h-x^{1/2})-(h+x)^{1/2}}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{(h-x^{1/2})^{2}-h-x}{h[(h-x^{1/2})-(h+x)^{1/2}]}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{h^{2}-2hx^{1/2}+x-h-x}{h[(h-x^{1/2})-(h+x)^{1/2}]}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{h[h-2x^{1/2}-1]}{h[(h-x^{1/2})-(h+x)^{1/2}]}\)
\(\displaystyle = \ \lim_{h\to0} \ \frac{h-2x^{1/2}-1}{(h-x^{1/2})-(h+x)^{1/2}} \ = \ \frac{-2x^{1/2}-1}{-x^{1/2}-x^{1/2}} \ = \ \frac{2x^{1/2}+1}{2x^{1/2}}, \ QED\)
\(\displaystyle Now, \ domains \ and \ ranges.\)
\(\displaystyle f(x) \ = \ x+x^{1/2}, hence \ x \ \ge 0, \ therefore \ domain \ is \ [0,\infty) \ and \ range \ is \ [0,\infty).\)
\(\displaystyle f \ ' \ (x) \ = \ \frac{2 \sqrt x+1}{2 \sqrt x}, \ hence \ x \ > \ 0, \ so \ domain \ is \ (0,\infty) \ and \ range \ is \ (1,\infty)\)
\(\displaystyle Note: \ \lim_{x\to0^{+}}\frac{2 \sqrt x+1}{2\sqrt x} \ = \ \infty \ and \ \lim_{x\to\infty}\frac{2\sqrt x+1}{2\sqrt x} \ = \ 1.\)
\(\displaystyle See \ the \ graph \ below. \ Although \ not \ shown, \ x=0 \ is \ a \ vertical \ aysmptote \ and \ y=1 \ is \ a \ horizontal\)
\(\displaystyle \ aysmptote.\)
[attachment=0:mqqbw5pp]zzz.jpg[/attachment:mqqbw5pp]
