Calculus Helpp!!!!. Urgent ( review)

Question

1) Suppose f(x), g(x), and h(x) are continuous functions such that g(−1)=5, h(−1)=3 and limx→−1 ((f(x))² − 3g(x)/ 5h(x)))=14. If f(x) is an odd function with f(1) > 0, find f(−1).

I got 15 but it's wrong please let me know the answer for this

2) Find and describe each discontinuity of f(x).

 (If an answer does not exist, enter DNE. Enter your answers as a comma-separated list.)

Let f(x) =( ^{squareroot of ((x^2+3x)-2)/ if x ≠ 1} )

x-1

-1 ^{if x = 1}

find:

removable discontinuity

jump discontinuity

infinite discontinuity

Answer 1

1) Suppose f(x), g(x), and h(x) are continuous functions such that g(−1)=5, h(−1)=3 and lim_x→−1 ((f(x))² − 3g(x)/ 5h(x)))=14. If f(x) is an odd function with f(1) > 0, find f(−1).

If lim_x→−1 ((f(x))² − 3g(x))/5h(x) = 14

then ((f(–1))² − 3g(–1))/5h(–1) = ((f(–1))² − 3(5))/5(3) = ((f(–1))² − 15)/15 = 14

Multiplying both sides by 15 gives

f(–1))² − 15 = 15×14 = 210

Adding 15 to both sides gives

(f(–1))² = 210 + 15 = 225

Taking the square root to both sides gives

f(–1) = ±√(225) = ±15

Now here's the key point

If f(x) is an odd function with f(1) > 0, then f(−1) < 0.

So f(–1) = –15

2) Assuming I'm interpreting the piecewise function correctly as

f(x) = (√(x² + 3x – 2))/(x –1) {x ≠ 1}; f(x) = –1 {x = 1}

discontinuities:

a. f(x) →∞ as x → 1 from the right; this is an infinite discontinuity

b. √(x² + 3x – 2) is undefined for ½(–3 –√17) < x < ½(–3 +√17) because x² + 3x – 2 is negative between

½(–3 –√17) and ½(–3 +√17).

This seems like a discontinuity to me, but I I don't know what-type/what-it's-called.

c. I'm not seeing either a removable discontinuity or a jump discontinuity.