Question
An extremum value of the function $f(x)=$ $\left(\sin ^{-1} x\right)^{3}+\left(\cos ^{-1} x\right)^{3}(-1<x<1)$ is(a) $7 \pi^{3} / 8$(b) $\pi^{3} / 8$(c) $\pi^{3} / 32$(d) $\pi^{3} / 16$
Step 1
We need to find the extremum of this function. Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 66 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The function $f(x)=\sin x \cos ^{2} x$ has extremum at (a) $x=\pi / 2$ (b) $x=\cos ^{-1}(v \sqrt{3})$ (c) $x=\cos ^{-1}(-\sqrt{23})$ (d) $x=\cos ^{-1}(-\sqrt{2 / 3})$
Let $f(x)=\left\{\begin{array}{lr}x^{3}+x^{2}-10 x, & -1 \leq x<0 \\ \cos x, & 0 \leq x<\pi / 2 \\ 1+\sin x, & \pi / 2 \leq x \leq \pi\end{array}\right.$ Then $\mathrm{f}(\mathrm{x})$ has (A) a local minimum at $x=\pi / 2$ (B) a global maximum at $\mathrm{x}=-1$ (C) an absolute minimum at $\mathrm{x}=-1$ (D) an absolute maximum at $\mathrm{x}=\pi$
Let $f(x)=\cos \pi x+10 x+3 x^{2}+x^{3},-2 \leq x \leq 3$. The absolute minimum value of $f(x)$ is (A) 0 (B) $-15$ (C) $3-2 \pi$ (D) None of these
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD