Question Number 12359 by sin (x) last updated on 20/Apr/17
Answered by ajfour last updated on 20/Apr/17
![for given range of a and b, (a^4 +b^3 )_(max) = 1−(1/8)= (7/8) (a^4 +b^3 )_(min) = 0+(−8+h )=−8+h so (c) → (−8,(7/8)] .](Q12360.png)
$${for}\:{given}\:{range}\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}, \\ $$$$\left(\boldsymbol{{a}}^{\mathrm{4}} +\boldsymbol{{b}}^{\mathrm{3}} \right)_{\boldsymbol{{max}}} \:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}=\:\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\left(\boldsymbol{{a}}^{\mathrm{4}} +\boldsymbol{{b}}^{\mathrm{3}} \right)_{\boldsymbol{{min}}} =\:\mathrm{0}+\left(−\mathrm{8}+{h}\:\right)=−\mathrm{8}+{h} \\ $$$$\boldsymbol{{so}}\:\left(\boldsymbol{{c}}\right)\:\rightarrow\:\left(−\mathrm{8},\frac{\mathrm{7}}{\mathrm{8}}\right]\:. \\ $$