We have f(x)=4x3−7,x∈R. fis one-one. Let x1,x2∈R and f(x1)=f(x2). ⇒4x13−7=4x23−7⇒4x13=4x2‌3 ⇒‌‌x13=x23⇒x13−x23=0 ⇒(x1−x2)(x12+x1x2+x22)=0 ⇒(x1−x2)[(x1+‌
x2
2
)2+‌
3x22
4
]=0 ⇒x1−x2=0, because the other factor is nonzero. ⇒x1=x2∴f is one-one. f is onto. Let k∈R any real number. f(x)=k⇒4x3−7=k⇒x=(‌
k+7
4
)1∕3 Now(‌
k+7
4
)1∕3∈R, because k∈R and f[(‌
k+7
4
)1∕3]=4[(‌
k+7
4
)1∕3]3−7 =4(‌
k+7
4
)−7=k ∴‌‌k is the image of (‌
k+7
4
)1∕3 ∴‌‌f is onto. ∴‌‌f is a bijective function.