To solve this problem, let's proceed step by step:
### Step 1: Find the First Derivative of [tex]\( f(x) \)[/tex]
Given the function:
[tex]\[ f(x) = 2.6 + 5.2x - 2.1x^2 \][/tex]
Take the first derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (2.6 + 5.2x - 2.1x^2) \][/tex]
[tex]\[ f'(x) = 5.2 - 4.2x \][/tex]
### Step 2: Find the Critical Numbers
Critical numbers occur where the first derivative is zero or undefined. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 5.2 - 4.2x = 0 \][/tex]
[tex]\[ 4.2x = 5.2 \][/tex]
[tex]\[ x = \frac{5.2}{4.2} \][/tex]
Simplify the fraction:
[tex]\[ x = \frac{26}{21} \][/tex]
[tex]\[ x = \frac{13}{10} \][/tex]
[tex]\[ x = 1.238 \quad (approx.) \][/tex]
Thus, the critical number is:
[tex]\[ \boxed{\frac{13}{10}} \][/tex]
### Step 3: Determine the Open Intervals Where the Function is Increasing or Decreasing
To determine intervals of increase and decrease, we need to analyze the sign of the first derivative [tex]\( f'(x) \)[/tex] around the critical number.
### Analyzing [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = 5.2 - 4.2x \][/tex]
#### For [tex]\( x < \frac{13}{10} \)[/tex]:
Pick a test point, like [tex]\( x = 0 \)[/tex].
[tex]\[ f'(0) = 5.2 - 4.2(0) = 5.2 \][/tex] (Positive)
Hence, [tex]\( f(x) \)[/tex] is increasing on the interval [tex]\( (-\infty, \frac{13}{10}) \)[/tex].
#### For [tex]\( x > \frac{13}{10} \)[/tex]:
Pick a test point, like [tex]\( x = 2 \)[/tex].
[tex]\[ f'(2) = 5.2 - 4.2(2) = 5.2 - 8.4 = -3.2 \][/tex] (Negative)
Hence, [tex]\( f(x) \)[/tex] is decreasing on the interval [tex]\( (\frac{13}{10}, \infty) \)[/tex].
### Summary
(a) The critical number is [tex]\( \frac{13}{10} \)[/tex] or [tex]\( 1.3 \)[/tex]. So, the correct choice is:
[tex]\[ \boxed{\frac{13}{10}} \][/tex]
(b) The function is increasing on the open interval:
[tex]\[ (-\infty, \frac{13}{10}) \][/tex]
(c) The function is decreasing on the open interval:
[tex]\[ (\frac{13}{10}, \infty) \][/tex]
