Can someone explain to me the steps how they got 2 √4^2+2x-3-2 from this question…

1 answer:

6 0

Step-by-step explanation:

with your square root symbol I never know what is inside the square root and what is possibly outside.

so, I can only guess and see what comes close.

f(x) = 2x² + x - 1

g(x) = sqrt(2x - 1) ??? is that so ?

h(x) = -2

2g(f(x)) + h(x)

g(f(x)) means that the whole f(x) expression is used as x in g(x).

the whole combined function is then

2×sqrt(2(2x² + x - 1) - 1) - 2

2×sqrt(4x² + 2x - 2 - 1) - 2

2×sqrt(4x² + 2x - 3) - 2

and if I am not mistaken, then this is the solution you mentioned at the beginning (if I try to read between the typos and missing info).

this is how people get to this.

do you understand it now ? or is there still something unclear ?

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tamaranim1 [39]

Hello!

All you have to is isolate the X variable :)

X + 25 = 45
X = 45 - 25

X = 20

I hope this helps :))

5 0

3 years ago

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Calculus

Differentiation

Basic Power Rule:

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$