Question

The function f(x)=x^9+3x^7+64 is increasing on

Answer 1

Answer:

f(x)=x9+3x7+64Differentiate with respect to xf'(x)=9x8+21x6As both powers are even powers (8,6), hencef'(x)≥0 for x∈RHence f(x) is increasing in x∈(−∞,∞

Step-by-step explanation:

Answer 2

Answer:

f(x) is ≥ 0 ∀x∈R

Step-by-step explanation:

the function $f(x) =$ $x^{9}+3x^{7}+64$

where x ∈ real numbes

$f'(x)=$ $9x^{8}+21x^{6}$

∵ $9x^{8}$ has a even power, it will alwats be ≥ 0 ∀$x$∈R

siilarly

$21x^{6}$≥ 0 ∀$x$∈R

since both numbers are ≥ 0, their sum wil be ≥ 0

so $f'(x)$ is ≥ 0 ∀$x$∈R

$f(x)=$ is increasing ∀$x$∈R