To determine whether the function [tex]\( f(x) = -5x^2 - 2x + 9 \)[/tex] has a maximum or minimum value, we follow these steps:
1. Find the first derivative of the function:
The first derivative [tex]\( f'(x) \)[/tex] is found by differentiating [tex]\( f(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx}(-5x^2 - 2x + 9) = -10x - 2
\][/tex]
2. Find the critical points:
Critical points occur where [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[
-10x - 2 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
-10x = 2 \quad \Rightarrow \quad x = -\frac{2}{10} = -\frac{1}{5}
\][/tex]
3. Determine the nature of the critical points using the second derivative:
We take the second derivative of [tex]\( f(x) \)[/tex]:
[tex]\[
f''(x) = \frac{d}{dx}(-10x - 2) = -10
\][/tex]
Evaluate the second derivative at the critical point [tex]\( x = -\frac{1}{5} \)[/tex]:
[tex]\[
f''\left( -\frac{1}{5} \right) = -10
\][/tex]
Since [tex]\( f''\left( -\frac{1}{5} \right) < 0 \)[/tex], the concavity at [tex]\( x = -\frac{1}{5} \)[/tex] is downward, indicating a local maximum at this critical point.
4. Find the value of the function at the critical point:
Substitute [tex]\( x = -\frac{1}{5} \)[/tex] back into the original function [tex]\( f(x) \)[/tex] to find the corresponding value:
[tex]\[
f\left( -\frac{1}{5} \right) = -5 \left( -\frac{1}{5} \right)^2 - 2 \left( -\frac{1}{5} \right) + 9
\][/tex]
Simplify the expression:
[tex]\[
= -5 \left( \frac{1}{25} \right) + \frac{2}{5} + 9
\][/tex]
[tex]\[
= -\frac{5}{25} + \frac{2}{5} + 9
\][/tex]
[tex]\[
= -\frac{1}{5} + \frac{2}{5} + 9
\][/tex]
[tex]\[
= \frac{1}{5} + 9
\][/tex]
[tex]\[
= \frac{1}{5} + \frac{45}{5}
\][/tex]
[tex]\[
= \frac{46}{5}
\][/tex]
Therefore, the function [tex]\( f(x) \)[/tex] has a maximum value and the corresponding value is [tex]\( \frac{46}{5} \)[/tex]. So, the correct choice is:
B. The function has a maximum value and the corresponding value is [tex]\(\frac{46}{5}\)[/tex].
