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erastovalidia [21]

3 years ago

11

What is the value of x when f(x)=7? the function or whatever is f (x) =x^3 - 1

2 answers:

mamaluj [8]3 years ago

8 0

F(7)= 342...........

pashok25 [27]3 years ago

4 0

Answer:

f ( 7 ) = 342

Step-by-step explanation:

Step 1:

f ( x ) = x³ - 1 Function

Step 2:

f ( 7 ) = 7³ - 1 Input x value

Step 3:

f ( 7 ) = 343 - 1 Exponents

Answer:

f ( 7 ) = 342 Subtract

Hope This Helps :)

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3 years ago

Find the derivative of f(x) = 12x^2 + 8x at x = 9.

zvonat [6]

Answer:

224

Step-by-step explanation:

We will need the following rules for derivative:

$(f+g)'=f'+g'$ Sum rule.

$(cf)'=cf'$ Constant multiple rule.

$(x^n)'=nx^{n-1}$ Power rule.

$(x)'=1$ Slope of y=x is 1.

$f(x)=12x^2+8x$

$f'(x)=(12x^2+8x)'$

$f'(x)=(12x^2)'+(8x)'$ by sum rule.

$f'(x)=12(x^2)+8(x)'$ by constant multiple rule.

$f'(x)=12(2x)+8(1)$ by power rule.

$f'(x)=24x+8$

Now we need to find the derivative function evaluated at x=9.

$f'(9)=24(9)+8$

$f'(9)=216+8$

$f'(9)=224$

In case you wanted to use the formal definition of derivative:

$f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Or the formal definition evaluated at x=a:

$f'(a)=\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$

Let's use that a=9.

$f'(9)=\lim_{h \rightarrow 0} \frac{f(9+h)-f(9)}{h}$

We need to find f(9+h) and f(9):

$f(9+h)=12(9+h)^2+8(9+h)$

$f(9+h)=12(9+h)(9+h)+72+8h$

$f(9+h)=12(81+18h+h^2)+72+8h$

(used foil or the formula (x+a)(x+a)=x^2+2ax+a^2)

$f(9+h)=972+216h+12h^2+72+8h$

Combine like terms:

$f(9+h)=1044+224h+12h^2$

$f(9)=12(9)^2+8(9)$

$f(9)=12(81)+72$

$f(9)=972+72$

$f(9)=1044$

Ok now back to our definition:

$f'(9)=\lim_{h \rightarrow 0} \frac{f(9+h)-f(9)}{h}$

$f'(9)=\lim_{h \rightarrow 0} \frac{1044+224h+12h^2-1044}{h}$

Simplify by doing 1044-1044:

$f'(9)=\lim_{h \rightarrow 0} \frac{224h+12h^2}{h}$

Each term has a factor of h so divide top and bottom by h:

$f'(9)=\lim_{h \rightarrow 0} \frac{224+12h}{1}$

$f'(9)=\lim_{h \rightarrow 0}(224+12h)$

$f'(9)=224+12(0)$

$f'(9)=224+0$

$f'(9)=224$

8 0

3 years ago