Answer:
[tex]f\left(-3\right)=\frac{126}{125}[/tex]
Step-by-step explanation:
Given
[tex]f\left(x\right)\:=\:\left(5\right)^x+1[/tex]
Putting x = -3 to find f(-3)
[tex]f\left(-3\right)\:=\:\left(5\right)^{\left(-3\right)}+1[/tex]
as
[tex]5^{-3}=\frac{1}{125}[/tex] ∵ [tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}[/tex]
so
[tex]f\left(-3\right)=\frac{1}{125}+1[/tex]
[tex]\mathrm{Convert\:element\:to\:fraction}:\quad \:1=\frac{1\cdot \:125}{125}[/tex]
[tex]f\left(-3\right)=\frac{1\cdot \:\:125}{125}+\frac{1}{125}[/tex]
[tex]\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}[/tex]
[tex]f\left(-3\right)=\frac{1\cdot \:\:125+1}{125}[/tex]
[tex]f\left(-3\right)=\frac{126}{125}[/tex] ∵ [tex]1\cdot \:125+1=126[/tex]
Thus,
[tex]f\left(-3\right)=\frac{126}{125}[/tex]
