If the domain of the function
f(x)=10+3x−x21+x+∣x∣1 is (a, b), then (1+a)2+b2 is equal to:
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Step-by-Step Solution
Step 1: Determine the domain for the first term
For the first term, 10+3x−x21, the expression under the square root must be strictly positive. This means 10+3x−x2>0. We can rewrite this inequality as x2−3x−10<0. Factoring the quadratic expression, we get (x - 5)(x + 2) < 0. This inequality holds true when x is between the roots, so −2<x<5.
Step 2: Determine the domain for the second term
For the second term, x+∣x∣1, the expression under the square root must also be strictly positive. This means x + |x| > 0. We consider two cases for ∣x∣: If x≥0, then ∣x∣=x, so x+x>0⟹2x>0⟹x>0. If x<0, then ∣x∣=−x, so x+(−x)>0⟹0>0, which is false. Therefore, the condition x + |x| > 0 is satisfied only when x>0.
Step 3: Find the intersection of the domains
The domain of the entire function f(x) is the intersection of the domains of its individual terms. We found that for the first term, the domain is (−2,5), and for the second term, the domain is (0,∞). The intersection of these two intervals is (0,5).
Step 4: Identify 'a' and 'b' and calculate the final expression
From the combined domain (0,5), we can identify a=0 and b=5. Now, we substitute these values into the expression (1+a)2+b2. This gives us (1+0)2+52=12+52=1+25=26.