Question

Let $f(x) = 8x^3 - 15x^2 - 5$, $x \ge 1.5$. Find the value of $\frac{df}{dx}^{-1}$ at the point $x = 2004 = f(7)$. The value of $\frac{df}{dx}^{-1}$ at the point $x = 2004 = f(7)$ is $\boxed{}$. (Type a simplified fraction.)

Name: let fx 8x3 15x2 5 x ge 15 find the value of fracdfdx 1 at the point x 2004 f7 the value of fracdfdx 1 at the point x 2004 f7 is boxed type a simplified fraction 65182
Uploaded: 2022-07-05T11:08:23-08:00
Duration: 3 min 55 s
Channel: Maitreya E
Description: let fx 8x3 15x2 5 x ge 15 find the value of fracdfdx 1 at the point x 2004 f7 the value of fracdfdx 1 at the point x 2004 f7 is boxed type a simplified fraction 65182

          Let $f(x) = 8x^3 - 15x^2 - 5$, $x \ge 1.5$. Find the value of $\frac{df}{dx}^{-1}$ at the point $x = 2004 = f(7)$.
The value of $\frac{df}{dx}^{-1}$ at the point $x = 2004 = f(7)$ is $\boxed{}$.
(Type a simplified fraction.)

$let fx 8x3 15x2 5 x ge 15 find the value of fracdfdx 1 at the point x 2004 f7 the value of fracdfdx 1 at the point x 2004 f7 is boxed type a simplified fraction 65182$

Added by Jeremy S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Maitreya E

07/05/2022

Step 1

$f(x) = 8x^3 - 15x^2 - 5$ $\frac{df}{dx} = 24x^2 - 30x$ Show more…

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Let $f(x) = 8x^3 - 15x^2 - 5$, $x \ge 1.5$. Find the value of $\frac{df}{dx}^{-1}$ at the point $x = 2004 = f(7)$. The value of $\frac{df}{dx}^{-1}$ at the point $x = 2004 = f(7)$ is $\boxed{}$. (Type a simplified fraction.)