Economy, asked by 7TeeN, 10 months ago

Find the points of inflexion for
$f(x) = (sinx + cosx) {e}^{x} \: where \: \\ belongs \: to \: 0 \: to \: 2\pi \:$
Discuss the concavity and convexity.

Answers

Answered by MRsteveAustiN

Differentiating To Find concavity

$f(x) = (sinx + cosx) {e}^{x} \: where \: \\ belongs \: to \: 0 \: to \: 2\pi \:$

$f'(x) = (cosx - sinx) {e}^{x}+(sinx + cosx) {e}^{x}$

$f'(x) = (2cosx) {e}^{x}$

$f"(x) = (-2sinx) {e}^{x} +(2cosx) {e}^{x}$

$f''(x) = (cosx-sinx)2 {e}^{x}$

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Answered by Aɾꜱɦ

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f(x)=(sinx+cosx)e

where

belongsto0to2π

f'(x) = (cosx - sinx) {e}^{x}+(sinx + cosx) {e}^{x}f

′

(x)=(cosx−sinx)e

+(sinx+cosx)e

f'(x) = (2cosx) {e}^{x}f

′

(x)=(2cosx)e

f"(x) = (-2sinx) {e}^{x} +(2cosx) {e}^{x}f"(x)=(−2sinx)e

+(2cosx)e

f''(x) = (cosx-sinx)2 {e}^{x}f

′′

(x)=(cosx−sinx)2e

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Find the points of inflexion for Discuss the concavity and convexity.​

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Find the points of inflexion for
$f(x) = (sinx + cosx) {e}^{x} \: where \: \\ belongs \: to \: 0 \: to \: 2\pi \:$
Discuss the concavity and convexity.