Find the maximum value of $$ 10+x-4 x^{2} $$ and the value of $$ x $$ for which the maximum occuns

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UpStudy AI · Accepted Answer

To find the maximum value of the quadratic expression $$ f(x) = 10 + x - 4x^{2} $$ and the corresponding value of $$ x $$ where this maximum occurs, follow these steps:

### 1. Identify the Quadratic Components

The given quadratic function can be rewritten in the standard form:
$$ f(x) = -4x^{2} + x + 10 $$
Here, the coefficients are:
- $$ a = -4 $$ (the coefficient of $$ x^2 $$)
- $$ b = 1 $$ (the coefficient of $$ x $$)
- $$ c = 10 $$ (the constant term)

### 2. Determine the Vertex of the Parabola

Since the coefficient $$ a = -4 $$ is negative, the parabola opens downward, and the function has a maximum value at its vertex.

The formula to find the $$ x $$-coordinate of the vertex (where the maximum occurs) is:
$$ x = -\frac{b}{2a} $$

Plugging in the values:
$$ x = -\frac{1}{2(-4)} = \frac{1}{8} $$

### 3. Calculate the Maximum Value

To find the maximum value of $$ f(x) $$, substitute $$ x = \frac{1}{8} $$ back into the original function:
$$
\begin{align*}
f\left(\frac{1}{8}\right) &= 10 + \frac{1}{8} - 4\left(\frac{1}{8}\right)^{2} \
&= 10 + \frac{1}{8} - 4\left(\frac{1}{64}\right) \
&= 10 + \frac{1}{8} - \frac{1}{16} \
&= \frac{160}{16} + \frac{2}{16} - \frac{1}{16} \
&= \frac{161}{16} \
&= 10.0625
\end{align*}
$$

### **Conclusion**

- **Maximum Value:** $$ \frac{161}{16} $$ (which is equal to 10.0625)
- **Value of $$ x $$ at Maximum:** $$ \frac{1}{8} $$

### **Final Answer**

The expression attains its maximum value of 161⁄16 when x is 1⁄8.
The maximum value of $$ 10 + x - 4x^{2} $$ is $$ \frac{161}{16} $$ (10.0625) when $$ x = \frac{1}{8} $$.

Find the maximum value of and the value of for which the maximum occuns

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