If y= e(power x) + e(power -x)/2 when y =1
Do you want to know x when y is 1?
y never gets as low as 1, for any value of x.
When x = 0, y = 1.500
When x = -0.4, y = 1.4162
When x = -0.45, y = 1.4286
When x = -0.5, y = 1.4309
When x = -0.6, y = 1.4599
However, knowing how loose folks are with parentheses around here, it's possible she meant
(e^x - e^(-x))/2 = sinh(x)
so,
e^x - e^-x = 2
e^x - 2 - e^-x = 0
clearing the fraction,
e^2x - 2e^x - 1 = 0
this is just a quadratic in e^x
e^x = (2 ± √8)/2
= 1 ± √2
e^x cannot be negative, so we have
e^x = 1 + √2
x = ln(1+√2) = 0.88137
check: asinh(1) = 0.88137
To find the value of x when y is equal to 1 in the equation y = e^x + e^(-x)/2, we can set up the equation as follows:
1 = e^x + e^(-x)/2
Now, let's simplify the equation by multiplying through by 2 to get rid of the fraction:
2 = 2e^x + e^(-x)
Next, let's rearrange the terms to isolate the exponential terms:
2e^x - e^(-x) = 2
Now, let's substitute e^(-x) with 1/e^x since e^(-x) is the reciprocal of e^x:
2e^x - 1/e^x = 2
Commonly, we multiply through by e^x to get rid of the fraction term:
(2e^x)(e^x) - (1/e^x)(e^x) = 2(e^x)
2e^(2x) - 1 = 2e^x
Now, let's move all terms to one side to set the equation equal to zero:
2e^(2x) - 2e^x - 1 = 0
At this point, we have a quadratic equation in terms of e^x. We can let z = e^x, which allows us to rewrite the equation as:
2z^2 - 2z - 1 = 0
Now, we can use the quadratic formula to solve for z:
z = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values for a, b, and c:
z = (-(-2) ± √((-2)^2 - 4(2)(-1))) / (2(2))
Simplifying further:
z = (2 ± √(4 + 8)) / 4
z = (2 ± √12) / 4
Finally, we can simplify the expression for z by factoring out the square root of 4 from the numerator:
z = (2 ± 2√3) / 4
Now, let's simplify further by dividing both the numerator and denominator by 2:
z = (1 ± √3) / 2
Since z = e^x, we can write:
e^x = (1 ± √3) / 2
To solve for x, we can take the natural logarithm (ln) of both sides:
ln(e^x) = ln((1 ± √3) / 2)
x = ln((1 ± √3) / 2)
Hence, the value of x when y is equal to 1 is x = ln((1 ± √3) / 2).