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The functions [tex]g[/tex] and [tex]h[/tex] are given by [tex]g(x) = x + 1[/tex] and [tex]h(x) = x^2[/tex].

b) Find the values of [tex]x[/tex] for which [tex]h(g(x)) = 3x^2 + x - 5[/tex].

Sagot :

Certainly! Let's solve the problem step by step.

We are given two functions [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex]:
[tex]\[ g(x) = x + 1 \][/tex]
[tex]\[ h(x) = x^2 \][/tex]

We need to find the values of [tex]\( x \)[/tex] for which [tex]\( h(g(x)) = 3x^2 + x - 5 \)[/tex].

1. Find [tex]\( h(g(x)) \)[/tex]:

Start by substituting [tex]\( g(x) \)[/tex] into [tex]\( h(x) \)[/tex]:
[tex]\[ g(x) = x + 1 \][/tex]
[tex]\[ h(g(x)) = h(x + 1) \][/tex]

Since [tex]\( h(x) = x^2 \)[/tex], substitute [tex]\( x + 1 \)[/tex] into [tex]\( x \)[/tex] in [tex]\( h(x) \)[/tex]:
[tex]\[ h(x + 1) = (x + 1)^2 \][/tex]

2. Expand the expression:

Expand [tex]\( (x + 1)^2 \)[/tex]:
[tex]\[ (x + 1)^2 = x^2 + 2x + 1 \][/tex]

So, [tex]\( h(g(x)) = x^2 + 2x + 1 \)[/tex].

3. Set up the equation:

We are given that [tex]\( h(g(x)) = 3x^2 + x - 5 \)[/tex]. Substitute [tex]\( h(g(x)) \)[/tex] with [tex]\( x^2 + 2x + 1 \)[/tex]:
[tex]\[ x^2 + 2x + 1 = 3x^2 + x - 5 \][/tex]

4. Solve for [tex]\( x \)[/tex]:

Move all terms to one side of the equation to set it to zero:
[tex]\[ x^2 + 2x + 1 - 3x^2 - x + 5 = 0 \][/tex]
Simplify the equation:
[tex]\[ -2x^2 + x + 6 = 0 \][/tex]
To make it more standard, multiply through by -1:
[tex]\[ 2x^2 - x - 6 = 0 \][/tex]

5. Factorize the quadratic equation:

We need to factor the equation [tex]\( 2x^2 - x - 6 = 0 \)[/tex]. To do this, we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \)[/tex].

First, calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-1)^2 - 4(2)(-6) \][/tex]
[tex]\[ \Delta = 1 + 48 \][/tex]
[tex]\[ \Delta = 49 \][/tex]

The discriminant is positive, so we have two real solutions. Now substitute back into the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{49}}{2(2)} \][/tex]
[tex]\[ x = \frac{1 \pm 7}{4} \][/tex]

This gives us two solutions:
[tex]\[ x = \frac{1 + 7}{4} = \frac{8}{4} = 2 \][/tex]
[tex]\[ x = \frac{1 - 7}{4} = \frac{-6}{4} = -\frac{3}{2} \][/tex]

Therefore, the values of [tex]\( x \)[/tex] for which [tex]\( h(g(x)) = 3x^2 + x - 5 \)[/tex] are:
[tex]\[ x = -\frac{3}{2} \quad \text{and} \quad x = 2 \][/tex]