17 mark for review selected values of the twice - differentiable functions f and g and their derivatives are…
17 mark for review selected values of the twice - differentiable functions f and g and their derivatives are given in the table above. the value of lim(x→2) (x²f(x)-16)/(g(x)-2) is a -28 b -12 c 28 d nonexistent
Answer
Explanation:
Step1: Check form of limit
When (x = 2), the numerator (x^{2}f(x)-16=2^{2}f(2)-16=4\times4 - 16=0), and the denominator (g(x)-2=g(2)-2=2 - 2=0). So, we can apply L - H rule.
Step2: Differentiate numerator and denominator
By the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{2}) and (v = f(x)), the derivative of the numerator (u^\prime=2x), (v^\prime = f^\prime(x)), so ((x^{2}f(x))^\prime=2xf(x)+x^{2}f^\prime(x)). The derivative of the denominator is (g^\prime(x)). The limit becomes (\lim_{x\rightarrow2}\frac{2xf(x)+x^{2}f^\prime(x)}{g^\prime(x)}).
Step3: Substitute (x = 2)
Substitute (x = 2) into (\frac{2xf(x)+x^{2}f^\prime(x)}{g^\prime(x)}). We get (\frac{2\times2\times f(2)+2^{2}\times f^\prime(2)}{g^\prime(2)}).
Step4: Use table values
From the table, (f(2) = 4), (f^\prime(2)=3), (g^\prime(2)=1). Then (\frac{2\times2\times4 + 4\times3}{1}=\frac{16 + 12}{1}=28).
Answer:
C. 28