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blagie [28]
3 years ago
11

Can someone help me please ?

Mathematics
1 answer:
Katarina [22]3 years ago
3 0

Answer: you can always write a square root, cubic root, etc... As an exponent.

Step-by-step explanation:

A) y^1/2

B) (y+6)^1/5 - 3

Exercise 4

A) 49^1/2 = square root of 49

(Dont have the square symbol y my phone. Lol!)

Always remember... You can write:

A^x/y = y root of A^x

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Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)
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Answer:

f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3

First tangent line:

y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5

Second tangent line:

y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7


Notice: slope of -1 means that both f'(x_1), \ f'(x_2) are equal to -1, so f'(x_1)=-1 \ and \ f'(x_2)=-1


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