Evaluate g(1+h)=x^2
The problem provided is asking you to evaluate a function g at a point close to 1, which is expressed as \(1+h\), where \(h\) is a small increment added to 1. It involves substituting the expression \(1+h\) into the function g and simplifying it accordingly. The problem statement includes a formula or expression for g that has to be known (implied here as a function of x, and represented by g(x) = x^2 when evaluated at x = 1) for the evaluation to take place. The task is to carry out the necessary algebraic manipulations to obtain g(1+h) in terms of h, usually with the intent of understanding the behavior of g near the point where x equals 1.
$g \left(\right. 1 + h \left.\right) = x^{2}$
Answer
Solution:
Step 1:
Divide both sides of the equation $g(1 + h) = x^2$ by the term $1 + h$ to isolate $g$.
$$\frac{g(1 + h)}{1 + h} = \frac{x^2}{1 + h}$$
Step 2:
Proceed to simplify the equation by focusing on the left-hand side.
Step 2.1:
Eliminate the common factor $(1 + h)$ from the numerator and the denominator.
Step 2.1.1:
Remove the shared term $(1 + h)$.
$$\frac{g(\cancel{1 + h})}{\cancel{1 + h}} = \frac{x^2}{1 + h}$$
Step 2.1.2:
Simplify the left side to find $g$.
$$g = \frac{x^2}{1 + h}$$
Knowledge Notes:
To solve for $g$ in the equation $g(1 + h) = x^2$, we need to isolate $g$ on one side of the equation. This is achieved by dividing both sides of the equation by the same non-zero term, which in this case is $(1 + h)$. This process is based on the property of equality that states if you divide both sides of an equation by the same non-zero quantity, the two sides remain equal.
When simplifying the left side of the equation, we notice that $g(1 + h)$ divided by $(1 + h)$ will cancel out the common factor of $(1 + h)$, leaving us with just $g$. This is because any term divided by itself equals one, and any term times one is itself.
On the right side, we are left with the expression $\frac{x^2}{1 + h}$, which cannot be simplified further without additional information about $x$ and $h$. The final step is to equate $g$ to this expression, which gives us the value of $g$ in terms of $x$ and $h$.
