Lada la función $$ f(x)=5 x^{2}-3 x+1 $$, hallar: $$ \begin{array}{ll} ext {-a) } f(0) & ext { f) } \frac{f(x+h)-f(x)}{h} \ ext { - b) } f(-1) & ext { g) } f(-2) \ ext { d) } f\left(\frac{1}{x} ight) & ext { i) } \frac{f(x+1 / 2)}{f(x+h)-f(x)} \ ext { e) } f(x+h) & \end{array} $$

Question

Lada la función $$ f(x)=5 x^{2}-3 x+1 $$, hallar: $$ \begin{array}{ll}	ext {-a) } f(0) & 	ext { f) } \frac{f(x+h)-f(x)}{h} \ 	ext { - b) } f(-1) & 	ext { g) } f(-2) \ 	ext { d) } f\left(\frac{1}{x}ight) & 	ext { i) } \frac{f(x+1 / 2)}{f(x+h)-f(x)} \ 	ext { e) } f(x+h) & \end{array} $$

UpStudy AI · Accepted Answer

\[
\begin{array}{rcl}
\textbf{a) } f(0) &=& 5\,(0)^2 - 3\,(0) + 1 = 1. \[1mm]
\textbf{b) } f(-1) &=& 5\,(-1)^2 - 3\,(-1) + 1 = 5(1) + 3 + 1 = 9. \[2mm]
\textbf{d) } f\left(\frac{1}{x}\right) &=& 5\left(\frac{1}{x}\right)^2 - 3\left(\frac{1}{x}\right) + 1 = \frac{5}{x^2} - \frac{3}{x} + 1. \[2mm]
\textbf{e) } f(x+h) &=& 5\,(x+h)^2 - 3\,(x+h) + 1 \
&=& 5\,(x^2+2xh+h^2) - 3x - 3h +1 \
&=& 5x^2 + 10xh + 5h^2 - 3x - 3h + 1. \[2mm]
\textbf{f) } \frac{f(x+h)-f(x)}{h} &=& \displaystyle \frac{\left[5(x+h)^2-3(x+h)+1\right] - \left[5x^2-3x+1\right]}{h}. \[2mm]
\text{Primero, se calcula } f(x+h)-f(x): \[2mm]
&& f(x+h)-f(x)= \Bigl(5x^2+10xh+5h^2-3x-3h+1\Bigr) - \Bigl(5x^2-3x+1\Bigr) \[2mm]
&& \quad = 10xh + 5h^2 - 3h. \[2mm]
\text{Por lo que:} \[2mm]
&& \frac{f(x+h)-f(x)}{h} = \frac{h(10x+5h-3)}{h} = 10x+5h-3. \[2mm]
\textbf{g) } f(-2) &=& 5\,(-2)^2 - 3\,(-2) + 1 = 5(4) + 6 + 1 = 27. $$2mm]
\textbf{i) } \frac{f(x+\frac{1}{2})}{f(x+h)-f(x)} &=& \displaystyle \frac{f\left(x+\frac{1}{2}\right)}{h(10x+5h-3)}. \[2mm]
\text{Calculamos } f\left(x+\frac{1}{2}\right): \[2mm]
&& f\left(x+\frac{1}{2}\right)= 5\left(x+\frac{1}{2}\right)^2 - 3\left(x+\frac{1}{2}\right)+1. \[2mm]
\text{Se expande:} \[2mm]
&& \left(x+\frac{1}{2}\right)^2 = x^2+x+\frac{1}{4}, \[2mm]
\text{entonces:} \[2mm]
&& f\left(x+\frac{1}{2}\right)= 5\left(x^2+x+\frac{1}{4}\right)- 3x - \frac{3}{2} + 1 \[2mm]
&& \quad = 5x^2+5x+\frac{5}{4}-3x-\frac{3}{2}+1 \[2mm]
&& \quad = 5x^2+2x + \left(\frac{5}{4} - \frac{3}{2}+1\right). \[2mm]
\text{Se unifican fracciones (usando denominador }4\text{):} \[2mm]
&& \frac{5}{4} - \frac{6}{4} + \frac{4}{4} = \frac{3}{4}. \[2mm]
\text{Así,} \[2mm]
&& f\left(x+\frac{1}{2}\right)= 5x^2+2x+\frac{3}{4}. \[2mm]
\text{Finalmente, la expresión del inciso i es:} \[2mm]
&& \frac{f(x+\frac{1}{2})}{f(x+h)-f(x)} = \frac{5x^2+2x+\frac{3}{4}}{h(10x+5h-3)}.
\end{array}
$$
$$ \begin{array}{rcl} \textbf{a) } f(0) &=& 1. \[1mm] \textbf{b) } f(-1) &=& 9. \[2mm] \textbf{d) } f\left(\frac{1}{x}\right) &=& \frac{5}{x^2} - \frac{3}{x} + 1. \[2mm] \textbf{e) } f(x+h) &=& 5x^2 + 10xh + 5h^2 - 3x - 3h + 1. \[2mm] \textbf{f) } \frac{f(x+h)-f(x)}{h} &=& 10x + 5h - 3. \[2mm] \textbf{g) } f(-2) &=& 27. \[2mm] \textbf{i) } \frac{f(x+\frac{1}{2})}{f(x+h)-f(x)} &=& \frac{5x^2+2x+\frac{3}{4}}{h(10x+5h-3)}. \end{array} $$

Lada la función \( f(x)=5 x^{2}-3 x+1 \), hallar: \( \begin{array}{ll}\text {-a) } f(0) & \text { f) } \frac{f(x+h)-f(x)}{h} \\ \text { - b) } f(-1) & \text { g) } f(-2) \\ \text { d) } f\left(\frac{1}{x}\right) & \text { i) } \frac{f(x+1 / 2)}{f(x+h)-f(x)} \\ \text { e) } f(x+h) & \end{array} \)

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