Question

Let $f(x) = e^{-3x^2}$. Then $f(x)$ has a relative minimum at x = DNE a relative maximum at x = DNE and inflection points at x = $\frac{1}{2}$ and at x = DNE

Name: let fx e 3x2 then fx has a relative minimum at x dne a relative maximum at x dne and inflection points at x frac12 and at x dne 88602
Uploaded: 2020-06-07T11:03:35-08:00
Duration: 2 min 2 s
Channel: Rakesh Kumar Sharma
Description: let fx e 3x2 then fx has a relative minimum at x dne a relative maximum at x dne and inflection points at x frac12 and at x dne 88602

          Let $f(x) = e^{-3x^2}$.
Then $f(x)$ has a relative minimum at
x = DNE
a relative maximum at
x = DNE
and inflection points at
x = $\frac{1}{2}$
and at
x = DNE

$let fx e 3x2 then fx has a relative minimum at x dne a relative maximum at x dne and inflection points at x frac12 and at x dne 88602$

Added by Jeremy W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Rakesh Kumar Sharma

06/07/2020

Step 1

$f'(x) = -6xe^{-3x^2}$ Show more…

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Let $f(x) = e^{-3x^2}$. Then $f(x)$ has a relative minimum at x = DNE a relative maximum at x = DNE and inflection points at x = $\frac{1}{2}$ and at x = DNE