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allochka39001 [22]
2 years ago
8

13. f(x) = -x + 4x^3 -2

Mathematics
1 answer:
Oksi-84 [34.3K]2 years ago
7 0

Answer:

x = 5

Step-by-step explanation:

You might be interested in
Solve the following system of equations by substitution. Show all steps.
Alinara [238K]

Answer:

Given system of equations:

\begin{cases}f(x)=-x^2+2x+3\\g(x)=-2x+3\end{cases}

To solve by substitution, equate the equations and solve for x:

\begin{aligned}f(x) & = g(x)\\\implies -x^2+2x+3 & = -2x+3\\-x^2+4x & = 0\\x^2-4x & = 0\\x(x-4) & = 0\\\implies x & = 0\\\implies x-4 & = 0 \implies x=4\end{aligned}

Therefore, the x-values of the solution are x = 0 and x = 4.

To find the y-values of the solution, substitute the found values of x into the functions:

f(0)=-(0)^2+2(0)+3=3

g(0)=-2(0)+3=3

f(4)=-(4)^2+2(4)+3=-5

g(4)=-2(4)+3=-5

Therefore, the solutions to the given system of equations are:

(0, 3) and (4, -5)

4 0
2 years ago
Anyone ? Need help. Thanks
NNADVOKAT [17]

Answer:

B.\:\: g(x)=4(2^x)

Step-by-step explanation:

The given function is f(x)=2^x.

If an exponential function is of the form; F(x)=a(b^x), then 0 will vertically shrink the base function f(x)=b^x

and a\:>\:1 will vertically stretch the graph by a units.

The correct answer is B.

7 0
3 years ago
Need help with this
Artemon [7]

The answer is D (the last answer choice)

3 0
3 years ago
13. The initial value of a particular vehicle is $20,000. After one year, the vehicle is worth $17,000. If the
Dima020 [189]
I think it would be D that’s what I think
7 0
3 years ago
Let f(x)=x^3-x-1
alexira [117]
f(x) = x^3 - x - 1

To find the gradient of the tangent, we must first differentiate the function.

f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

y = -x + c

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

f(0) = (0)^3 - (0) - 1 = -1

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
6 0
3 years ago
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