Find out whether the following functions are one to one or not
(a) f(x)=3x+4 (b) f(x)=x²+4 (c) h(x)=13x⁵+5 (d) g(x)=x⁴+3 (e) p(x)=1/x+3 (f) f(x) =|2x+5|
a:f(x)=f(y)⇒3x+4=3y+4⇒x=y−yesb:f(1)=5f(−1)=5Noc:h(x)=h(y)⇒13x5+5=13y5+5⇒x=y−yesd:g(1)=4g(−1)=4Noe:p(x)=p(y)⇒1x+3=1y+3⇒x=y−yesf:f(0)=5f(−5)=5Noa:\\f\left( x \right) =f\left( y \right) \Rightarrow 3x+4=3y+4\Rightarrow x=y-yes\\b:\\f\left( 1 \right) =5\\f\left( -1 \right) =5\\No\\c:\\h\left( x \right) =h\left( y \right) \Rightarrow 13x^5+5=13y^5+5\Rightarrow x=y-yes\\d:\\g\left( 1 \right) =4\\g\left( -1 \right) =4\\No\\e:\\p\left( x \right) =p\left( y \right) \Rightarrow \frac{1}{x+3}=\frac{1}{y+3}\Rightarrow x=y-yes\\f:\\f\left( 0 \right) =5\\f\left( -5 \right) =5\\Noa:f(x)=f(y)⇒3x+4=3y+4⇒x=y−yesb:f(1)=5f(−1)=5Noc:h(x)=h(y)⇒13x5+5=13y5+5⇒x=y−yesd:g(1)=4g(−1)=4Noe:p(x)=p(y)⇒x+31=y+31⇒x=y−yesf:f(0)=5f(−5)=5No
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